Squares Under 7: A Digitastic Number Theory Puzzle

Hey guys! Today, we're diving deep into a super cool number theory problem that's as mind-bending as it is fascinating. We're talking about perfect squares – those numbers you get when you multiply an integer by itself, like 4 (2x2), 9 (3x3), or 16 (4x4). But not just any perfect squares will do for this puzzle. We've got some strict criteria, and I'm stoked to break them down with you.

Our main quest is to find perfect squares that satisfy two pretty wild conditions. First off, every single digit in these numbers has to be strictly less than 7. This means you won't find any 7s, 8s, or 9s in our target numbers. We're talking digits from the set {0, 1, 2, 3, 4, 5, 6}. Think of it as a number system with a limited palette! The second condition is where things get really spicy: if you take each digit of these perfect squares and add 3 to it, the resulting number must also be a perfect square. It's like a double-whammy of perfection we're hunting for.

This problem, right at the intersection of number theory and elementary number theory, is a fantastic exercise in logical deduction and computational exploration. We're not just looking for a few examples; we're aiming to classify all such numbers. This means we need a systematic way to find them, prove they are the only ones, and understand the underlying patterns. So, grab your thinking caps, because we're about to embark on a mathematical adventure!

The Digit Constraint: A Palate of Possibilities

Let's kick things off by really digging into the first condition: every digit of the perfect square N must be strictly less than 7. This is a huge constraint, guys, and it immediately narrows down the possibilities significantly. When we talk about digits, we're referring to the individual symbols used in the base-10 representation of a number. So, for our target number $N$, each digit $d$ must satisfy $0 ":{255,0,0}" $d < 7$. This means the only allowed digits are 0, 1, 2, 3, 4, 5, and 6. If we see a 7, 8, or 9 anywhere in the number, it's immediately disqualified. This is like trying to paint a masterpiece using only a limited set of colors – it forces creativity and strategic placement.

Consider the implications of this rule. Large numbers, especially, are less likely to meet this criterion. As numbers grow, the probability of encountering higher digits increases. For instance, numbers ending in 00 are common perfect squares (like 100, 400, 900), and 0 is allowed. Numbers ending in 1, 4, 5, 6, and 9 are also possible as perfect squares. However, the constraint isn't just about the last digit; it's about every digit. This means a number like 144 is allowed under this rule (digits 1 and 4 are less than 7), but 169 is not (because of the 9). The number 49 is also out because of the 9. This initial filter is crucial because it prunes a vast number of potential candidates right from the start. It forces us to think about numbers that are composed only of these smaller digits. We're essentially working within a restricted set of numerical building blocks. This constraint is not just a simple rule; it's a fundamental characteristic that shapes the nature of the numbers we are seeking. We are searching for perfect squares that are, in a sense, numerically 'modest' in their composition, avoiding the 'larger' digits that might seem more common in the grand scheme of numbers.

Furthermore, this digit constraint directly impacts the magnitude of the perfect squares we can consider. If we think about squares like $30^2 = 900$, the digit 9 immediately disqualifies it. $20^2 = 400$ is fine. $25^2 = 625$ is fine. $26^2 = 676$ is fine. $27^2 = 729$ is not fine because of the 7 and 9. $31^2 = 961$ is not fine. $32^2 = 1024$ is not fine because of the 4. Wait, $32^2=1024$? No, 1024 is ok since 1,0,2,4 are all less than 7. My mistake there! $32^2 = 1024$. The digits are 1, 0, 2, 4. All are strictly less than 7. So 1024 passes the first test! This highlights how important it is to be meticulous. We need to check every digit. Let's re-evaluate. $30^2 = 900$ - disqualified (9). $31^2 = 961$ - disqualified (9). $32^2 = 1024$ - allowed (1, 0, 2, 4 < 7). $33^2 = 1089$ - disqualified (8, 9). $34^2 = 1156$ - allowed (1, 1, 5, 6 < 7). $35^2 = 1225$ - allowed (1, 2, 2, 5 < 7). $36^2 = 1296$ - disqualified (9). $37^2 = 1369$ - disqualified (9). $38^2 = 1444$ - allowed (1, 4, 4, 4 < 7). $39^2 = 1521$ - allowed (1, 5, 2, 1 < 7). $40^2 = 1600$ - allowed (1, 6, 0, 0 < 7). $41^2 = 1681$ - disqualified (8). $42^2 = 1764$ - disqualified (7). $43^2 = 1849$ - disqualified (8, 9). $44^2 = 1936$ - disqualified (9). $45^2 = 2025$ - allowed (2, 0, 2, 5 < 7). $46^2 = 2116$ - allowed (2, 1, 1, 6 < 7). $47^2 = 2209$ - disqualified (9). $48^2 = 2304$ - allowed (2, 3, 0, 4 < 7). $49^2 = 2401$ - allowed (2, 4, 0, 1 < 7). $50^2 = 2500$ - allowed (2, 5, 0, 0 < 7). $51^2 = 2601$ - allowed (2, 6, 0, 1 < 7). $52^2 = 2704$ - disqualified (7). $53^2 = 2809$ - disqualified (8, 9). $54^2 = 2916$ - disqualified (9). $55^2 = 3025$ - allowed (3, 0, 2, 5 < 7). $56^2 = 3136$ - allowed (3, 1, 3, 6 < 7). $57^2 = 3249$ - disqualified (9). $58^2 = 3364$ - allowed (3, 3, 6, 4 < 7). $59^2 = 3481$ - disqualified (8). $60^2 = 3600$ - allowed (3, 6, 0, 0 < 7). $61^2 = 3721$ - disqualified (7). $62^2 = 3844$ - disqualified (8). $63^2 = 3969$ - disqualified (9). $64^2 = 4096$ - disqualified (9). $65^2 = 4225$ - allowed (4, 2, 2, 5 < 7). $66^2 = 4356$ - allowed (4, 3, 5, 6 < 7). $67^2 = 4489$ - disqualified (8, 9). $68^2 = 4624$ - allowed (4, 6, 2, 4 < 7). $69^2 = 4761$ - disqualified (7). $70^2 = 4900$ - disqualified (9). $71^2 = 5041$ - allowed (5, 0, 4, 1 < 7). $72^2 = 5184$ - disqualified (8). $73^2 = 5329$ - disqualified (9). $74^2 = 5476$ - disqualified (7). $75^2 = 5625$ - allowed (5, 6, 2, 5 < 7). $76^2 = 5776$ - disqualified (7). $77^2 = 5929$ - disqualified (9). $78^2 = 6084$ - disqualified (8). $79^2 = 6241$ - allowed (6, 2, 4, 1 < 7). $80^2 = 6400$ - allowed (6, 4, 0, 0 < 7). $81^2 = 6561$ - allowed (6, 5, 6, 1 < 7). $82^2 = 6724$ - disqualified (7). $83^2 = 6889$ - disqualified (8, 9). $84^2 = 7056$ - disqualified (7, 8, 9). $85^2 = 7225$ - disqualified (7, 8, 9). $86^2 = 7396$ - disqualified (7, 8, 9). $87^2 = 7569$ - disqualified (7, 8, 9). $88^2 = 7744$ - disqualified (7, 8, 9). $89^2 = 7921$ - disqualified (7, 8, 9). $90^2 = 8100$ - disqualified (8). $91^2 = 8281$ - disqualified (8). $92^2 = 8464$ - disqualified (8). $93^2 = 8649$ - disqualified (8, 9). $94^2 = 8836$ - disqualified (8). $95^2 = 9025$ - disqualified (9). $96^2 = 9216$ - disqualified (9). $97^2 = 9409$ - disqualified (9). $98^2 = 9604$ - disqualified (9). $99^2 = 9801$ - disqualified (9). $100^2 = 10000$ - allowed. $101^2 = 10201$ - allowed. $102^2 = 10404$ - allowed. $103^2 = 10609$ - disqualified (9). $104^2 = 10816$ - disqualified (8). $105^2 = 11025$ - allowed. $106^2 = 11236$ - allowed. $107^2 = 11449$ - disqualified (9). $108^2 = 11664$ - allowed. $109^2 = 11881$ - disqualified (8). $110^2 = 12100$ - allowed. $111^2 = 12321$ - allowed. $112^2 = 12544$ - allowed. $113^2 = 12769$ - disqualified (7, 9). $114^2 = 12996$ - disqualified (9). $115^2 = 13225$ - allowed. $116^2 = 13456$ - allowed. $117^2 = 13689$ - disqualified (8, 9). $118^2 = 13924$ - disqualified (9). $119^2 = 14161$ - allowed. $120^2 = 14400$ - allowed. $121^2 = 14641$ - allowed. $122^2 = 14884$ - disqualified (8). $123^2 = 15129$ - disqualified (9). $124^2 = 15376$ - disqualified (7). $125^2 = 15625$ - allowed. $126^2 = 15876$ - disqualified (8, 7). $127^2 = 16129$ - disqualified (9). $128^2 = 16384$ - disqualified (8). $129^2 = 16641$ - allowed. $130^2 = 16900$ - disqualified (9). $131^2 = 17161$ - disqualified (7). $132^2 = 17424$ - disqualified (7). $133^2 = 17689$ - disqualified (7, 8, 9). $134^2 = 17956$ - disqualified (7, 9). $135^2 = 18225$ - disqualified (8). $136^2 = 18496$ - disqualified (8, 9). $137^2 = 18769$ - disqualified (8, 7, 9). $138^2 = 19044$ - disqualified (9). $139^2 = 19321$ - disqualified (9). $140^2 = 19600$ - disqualified (9). $141^2 = 19881$ - disqualified (9, 8). $142^2 = 20164$ - allowed. $143^2 = 20449$ - disqualified (9). $144^2 = 20736$ - disqualified (7). $145^2 = 21025$ - allowed. $146^2 = 21316$ - allowed. $147^2 = 21609$ - disqualified (9). $148^2 = 21904$ - disqualified (9). $149^2 = 22201$ - allowed. $150^2 = 22500$ - allowed. $151^2 = 22801$ - disqualified (8). $152^2 = 23104$ - allowed. $153^2 = 23409$ - disqualified (9). $154^2 = 23716$ - disqualified (7). $155^2 = 24025$ - allowed. $156^2 = 24336$ - allowed. $157^2 = 24649$ - disqualified (9). $158^2 = 24964$ - disqualified (9). $159^2 = 25281$ - disqualified (8). $160^2 = 25600$ - allowed. $161^2 = 25921$ - disqualified (9). $162^2 = 26244$ - allowed. $163^2 = 26569$ - disqualified (9). $164^2 = 26896$ - disqualified (8, 9). $165^2 = 27225$ - disqualified (7). $166^2 = 27556$ - disqualified (7). $167^2 = 27889$ - disqualified (7, 8, 9). $168^2 = 28224$ - disqualified (8). $169^2 = 28561$ - disqualified (8). $170^2 = 28900$ - disqualified (8, 9). $171^2 = 29241$ - disqualified (9). $172^2 = 29584$ - disqualified (9, 8). $173^2 = 29929$ - disqualified (9). $174^2 = 30276$ - disqualified (7). $175^2 = 30625$ - allowed. $176^2 = 30976$ - disqualified (9, 7). $177^2 = 31329$ - disqualified (9). $178^2 = 31684$ - disqualified (8). $179^2 = 32041$ - allowed. $180^2 = 32400$ - allowed. $181^2 = 32761$ - disqualified (7). $182^2 = 33124$ - allowed. $183^2 = 33489$ - disqualified (8, 9). $184^2 = 33856$ - disqualified (8). $185^2 = 34225$ - allowed. $186^2 = 34596$ - disqualified (9). $187^2 = 34969$ - disqualified (9). $188^2 = 35344$ - allowed. $189^2 = 35721$ - disqualified (7). $190^2 = 36100$ - allowed. $191^2 = 36481$ - disqualified (8). $192^2 = 36864$ - allowed. $193^2 = 37249$ - disqualified (7, 9). $194^2 = 37636$ - disqualified (7). $195^2 = 38025$ - disqualified (8). $196^2 = 38416$ - disqualified (8). $197^2 = 38809$ - disqualified (8, 9). $198^2 = 39204$ - disqualified (9). $199^2 = 39601$ - disqualified (9). $200^2 = 40000$ - allowed. $201^2 = 40401$ - allowed. $202^2 = 40804$ - disqualified (8). $203^2 = 41209$ - disqualified (9). $204^2 = 41616$ - allowed. $205^2 = 42025$ - allowed. $206^2 = 42436$ - allowed. $207^2 = 42849$ - disqualified (8, 9). $208^2 = 43264$ - allowed. $209^2 = 43681$ - disqualified (8). $210^2 = 44100$ - allowed. $211^2 = 44521$ - allowed. $212^2 = 44944$ - disqualified (9). $213^2 = 45369$ - disqualified (9). $214^2 = 45796$ - disqualified (7, 9). $215^2 = 46225$ - allowed. $216^2 = 46656$ - allowed. $217^2 = 47089$ - disqualified (7, 8, 9). $218^2 = 47524$ - disqualified (7). $219^2 = 47961$ - disqualified (7, 9). $220^2 = 48400$ - disqualified (8). $221^2 = 48841$ - disqualified (8). $222^2 = 49284$ - disqualified (9, 8). $223^2 = 49729$ - disqualified (9, 7). $224^2 = 50176$ - disqualified (7). $225^2 = 50625$ - allowed. $226^2 = 51076$ - disqualified (7). $227^2 = 51529$ - disqualified (9). $228^2 = 51984$ - disqualified (9, 8). $229^2 = 52441$ - allowed. $230^2 = 52900$ - allowed. $231^2 = 53361$ - allowed. $232^2 = 53824$ - disqualified (8). $233^2 = 54289$ - disqualified (8, 9). $234^2 = 54756$ - disqualified (7). $235^2 = 55225$ - allowed. $236^2 = 55696$ - disqualified (9). $237^2 = 56169$ - disqualified (9). $238^2 = 56644$ - allowed. $239^2 = 57121$ - disqualified (7). $240^2 = 57600$ - disqualified (7). $241^2 = 58081$ - disqualified (8). $242^2 = 58564$ - disqualified (8). $243^2 = 59049$ - disqualified (9). $244^2 = 59536$ - disqualified (9). $245^2 = 60025$ - allowed. $246^2 = 60516$ - allowed. $247^2 = 61009$ - disqualified (9). $248^2 = 61504$ - allowed. $249^2 = 62001$ - allowed. $250^2 = 62500$ - allowed. $251^2 = 63001$ - allowed. $252^2 = 63504$ - allowed. $253^2 = 64009$ - disqualified (9). $254^2 = 64516$ - allowed. $255^2 = 65025$ - allowed. $256^2 = 65536$ - allowed. $257^2 = 66049$ - disqualified (9). $258^2 = 66564$ - allowed. $259^2 = 67081$ - disqualified (7). $260^2 = 67600$ - disqualified (7). $261^2 = 68121$ - disqualified (8). $262^2 = 68644$ - disqualified (8). $263^2 = 69169$ - disqualified (9). $264^2 = 69696$ - disqualified (9). $265^2 = 70225$ - disqualified (7). $266^2 = 70756$ - disqualified (7). $267^2 = 71289$ - disqualified (7, 8, 9). $268^2 = 71824$ - disqualified (7, 8). $269^2 = 72361$ - disqualified (7). $270^2 = 72900$ - disqualified (7, 9). $271^2 = 73441$ - disqualified (7). $272^2 = 73984$ - disqualified (7, 9, 8). $273^2 = 74529$ - disqualified (7, 9). $274^2 = 75076$ - disqualified (7). $275^2 = 75625$ - disqualified (7). $276^2 = 76176$ - disqualified (7). $277^2 = 76729$ - disqualified (7, 9). $278^2 = 77284$ - disqualified (7). $279^2 = 77841$ - disqualified (7). $280^2 = 78400$ - disqualified (8, 7). $281^2 = 78961$ - disqualified (8, 7, 9). $282^2 = 79524$ - disqualified (7, 9). $283^2 = 80089$ - disqualified (8, 9). $284^2 = 80656$ - disqualified (8). $285^2 = 81225$ - disqualified (8). $286^2 = 81796$ - disqualified (8, 9). $287^2 = 82369$ - disqualified (8, 9). $288^2 = 82944$ - disqualified (8, 9). $289^2 = 83521$ - disqualified (8). $290^2 = 84100$ - disqualified (8). $291^2 = 84681$ - disqualified (8). $292^2 = 85264$ - disqualified (8). $293^2 = 85849$ - disqualified (8, 9). $294^2 = 86436$ - disqualified (8). $295^2 = 87025$ - disqualified (8, 7). $296^2 = 87616$ - disqualified (8, 7). $297^2 = 88209$ - disqualified (8, 9). $298^2 = 88704$ - disqualified (8). $299^2 = 89401$ - disqualified (9, 8). $300^2 = 90000$ - disqualified (9). $301^2 = 90601$ - disqualified (9). $302^2 = 91204$ - disqualified (9). $303^2 = 91809$ - disqualified (9). $304^2 = 92416$ - disqualified (9). $305^2 = 93025$ - disqualified (9). $306^2 = 93636$ - disqualified (9). $307^2 = 94249$ - disqualified (9). $308^2 = 94864$ - disqualified (9). $309^2 = 95481$ - disqualified (9). $310^2 = 96100$ - disqualified (9). $311^2 = 96721$ - disqualified (9). $312^2 = 97344$ - disqualified (9, 7). $313^2 = 97969$ - disqualified (9). $314^2 = 98596$ - disqualified (9, 8). $315^2 = 99225$ - disqualified (9). $316^2 = 99856$ - disqualified (9). This exercise shows that the first condition drastically limits the search space. We're left with a manageable, albeit still substantial, list of candidates.

The Digit Transformation: A Square-ception!

The second condition is where the real magic, and the challenge, lies: adding 3 to each digit yields another square. This is what we call a 'digit transformation'. Let $N$ be a perfect square whose digits $d_k d_{k-1} ext{ extellipsis } d_1 d_0$ are all less than 7. The condition states that if we form a new number $N'$ by replacing each digit $d_i$ with $d_i' = d_i + 3$, then $N'$ must also be a perfect square. Remember, the digits $d_i$ are from {0, 1, 2, 3, 4, 5, 6}. Therefore, the transformed digits $d_i'$ will be from the set {3, 4, 5, 6, 7, 8, 9}.

Let's take an example. Suppose we have a number $N = 1024$. We already confirmed its digits (1, 0, 2, 4) are all less than 7. Now, let's apply the transformation: $1+3=4$, $0+3=3$, $2+3=5$, $4+3=7$. The new sequence of digits is 4, 3, 5, 7. The number formed is $N' = 4357$. Is 4357 a perfect square? Let's check. $\sqrt{4357} \approx 66.007$. No, it's not a perfect square. So, $N = 1024$ is not a solution.

What about $N = 1156$? Digits are 1, 1, 5, 6 (all < 7). Transform: $1+3=4$, $1+3=4$, $5+3=8$, $6+3=9$. The new number is $N' = 4489$. Is 4489 a perfect square? $\sqrt{4489} = 67$. Yes, it is! So, $N = 1156$ is a potential candidate! This is awesome!

Let's try another one: $N = 1444$. Digits are 1, 4, 4, 4 (all < 7). Transform: $1+3=4$, $4+3=7$, $4+3=7$, $4+3=7$. The new number is $N' = 4777$. Is 4777 a perfect square? $\sqrt{4777} \approx 69.11$. No, it's not. So $N = 1444$ is out.

Consider the transformed digits $d_i'$. They can be 3, 4, 5, 6, 7, 8, 9. For $N'$ to be a perfect square, its digits must follow the rules of squares. A key observation here is that the last digit of a perfect square can only be 0, 1, 4, 5, 6, or 9. This means that the last digit of $N$, $d_0$, when transformed to $d_0' = d_0 + 3$, must result in one of these allowed last digits for squares.

Let's analyze this:

• If $d_0 = 0$, $d_0' = 3$ (not a square's last digit).
• If $d_0 = 1$, $d_0' = 4$ (possible last digit).
• If $d_0 = 2$, $d_0' = 5$ (possible last digit).
• If $d_0 = 3$, $d_0' = 6$ (possible last digit).
• If $d_0 = 4$, $d_0' = 7$ (not a square's last digit).
• If $d_0 = 5$, $d_0' = 8$ (not a square's last digit).
• If $d_0 = 6$, $d_0' = 9$ (possible last digit).

So, the last digit $d_0$ of our original perfect square $N$ can only be 1, 2, 3, or 6. This is a powerful deduction! It immediately eliminates many candidates whose last digit isn't in this set.

Now let's think about the other digits $d_i$ (where $i > 0$). These transformed digits $d_i'$ can be {3, 4, 5, 6, 7, 8, 9}. We know that for $N'$ to be a perfect square, all its digits must be compatible with being part of a square. However, the restriction on $N$ is that its digits $d_i$ are strictly less than 7. This means $d_i e 7, 8, 9$. Consequently, the transformed digits $d_i' = d_i + 3$ can be 7, 8, or 9. For example, if $d_i = 4$, $d_i' = 7$. If $d_i = 5$, $d_i' = 8$. If $d_i = 6$, $d_i' = 9$.

Crucially, the digits of a perfect square cannot be 2, 3, 7, or 8. If any digit in $N'$ is 2, 3, 7, or 8, then $N'$ cannot be a perfect square. This means that for any digit $d_i$ in $N$ (other than possibly the last digit), the transformed digit $d_i' = d_i + 3$ cannot be 2, 3, 7, or 8.

Let's check:

• If $d_i = 0$, $d_i' = 3$. This is not allowed for a digit in $N'$ (unless $N'$ is a single digit number, which is not the case here since $N'$ is formed by transforming digits of $N$, and $N$ has at least one digit, leading to $N'$ having at least one digit). So, $d_i=0$ is problematic for digits other than the last one. This means if $d_i=0$ for $i>0$, then $d_i'=3$, which cannot be a digit of a perfect square $N'$. Wait, this is incorrect. The digits of $N'$ can be 3, 4, 5, 6, 7, 8, 9. However, the combination of digits matters. For example, $N'=49$ is a square ($7^2$), but its digits are 4 and 9. $N'=36$ is a square ($6^2$), digits 3 and 6. $N'=81$ is a square ($9^2$), digits 8 and 1. The rule is that no digit other than 0, 1, 4, 5, 6, 9 can appear in a perfect square. So, if $d_i' = 2, 3, 7, 8$, then $N'$ cannot be a perfect square.

Let's re-evaluate the transformed digits $d_i' = d_i + 3$ for $d_i less 7$.

• If $d_i = 0$, $d_i' = 3$. This digit '3' cannot appear in $N'$. So $d_i$ cannot be 0 for any $i>0$. Wait, this is wrong. A number like $1369 = 37^2$ has digits 3 and 6, which are allowed. A number like $4356 = 66^2$ has digits 4, 3, 5, 6, which are allowed. Ah, I confused the condition on $N$ with the condition on $N'$. $N'$ can have digits 3, 4, 5, 6, 7, 8, 9. The question is whether the number formed by these digits is a square.

Let's return to the analysis of $d_i'$ values:

• If $d_i = 0$, $d_i' = 3$. So, if $N$ contains a 0, the transformed number $N'$ will contain a 3. Can a perfect square contain the digit 3? Yes, e.g., $36 = 6^2$, $1369 = 37^2$, $4356 = 66^2$. So, $d_i=0$ is allowed for $N$ if $d_i'=3$ is allowed in $N'$.
• If $d_i = 1$, $d_i' = 4$. Allowed.
• If $d_i = 2$, $d_i' = 5$. Allowed.
• If $d_i = 3$, $d_i' = 6$. Allowed.
• If $d_i = 4$, $d_i' = 7$. Allowed.
• If $d_i = 5$, $d_i' = 8$. Allowed.
• If $d_i = 6$, $d_i' = 9$. Allowed.

The problem seems to be a combination of number theory properties and computational search. We've established that the last digit $d_0$ must be 1, 2, 3, or 6. Let's re-examine the digits of $N'$: {3, 4, 5, 6, 7, 8, 9}. Crucially, the last digit of $N'$ must be one of {0, 1, 4, 5, 6, 9}. Since $d_0' = d_0 + 3$, we already deduced $d_0 o d_0'$ must map to one of {4, 5, 6, 9}. This is consistent.

What about the other digits of $N'$? Let's consider potential digits of $N'$: {3, 4, 5, 6, 7, 8, 9}. Are there any restrictions on these digits within a square number? Yes, the digits in a perfect square cannot end with 2, 3, 7, or 8 if they appear in certain positions or combinations. However, the simple rule is that no perfect square ends in 2, 3, 7, or 8. This means if $d_i+3$ results in 2, 3, 7, or 8 for any digit $d_i$ in $N$, then $N'$ cannot be a square.

Let's check this again:

• $d_i = 0 ightarrow d_i' = 3$. So, if $N$ has a 0 digit, $N'$ will have a 3. $N'$ cannot contain the digit 3 if it's to be a perfect square. Wait, this is WRONG. $36 = 6^2$ contains 3. $1369 = 37^2$ contains 3. The property is that perfect squares cannot end in 3, 7, 8, 2. But they can contain these digits elsewhere. This makes the problem much harder!

Let's reconsider the implications of $d_i+3$ for $d_i eq d_0$. The possible values for $d_i+3$ are {3, 4, 5, 6, 7, 8, 9}.

If $d_i = 0$, $d_i+3=3$. So $N'$ contains a 3. Can $N'$ be a square? Yes, if it doesn't end in 3. Example: $N=1156$, $N'=4489$. $N$ has digits {1, 5, 6}. $N'$ has digits {4, 8, 9}. $N'$ is $67^2$. OK.

If $d_i = 1$, $d_i+3=4$. $N'$ contains a 4. Possible. If $d_i = 2$, $d_i+3=5$. $N'$ contains a 5. Possible. If $d_i = 3$, $d_i+3=6$. $N'$ contains a 6. Possible. If $d_i = 4$, $d_i+3=7$. $N'$ contains a 7. Possible. If $d_i = 5$, $d_i+3=8$. $N'$ contains an 8. Possible. If $d_i = 6$, $d_i+3=9$. $N'$ contains a 9. Possible.

The key insight might be the number of digits. If $N$ has $k+1$ digits, and none of them are 7, 8, 9, then $N$ is roughly between $10^k$ and $7 imes 10^k$. The transformed number $N'$ would have digits $d_i+3$. The smallest possible transformed digit is $0+3=3$, and the largest is $6+3=9$. So $N'$ will also have roughly the same number of digits as $N$.

Let $N = m^2$ and $N' = (m')^2$. The digits of $N$ are $d_i eq 7, 8, 9$. The digits of $N'$ are $d_i' = d_i+3$. We know $d_0 e 4, 5$ from the last digit analysis. So $d_0 o d_0'$ maps to {3, 6, 9}. This means the last digit of $N$ can only be 1, 2, 3, 6.

Let's test candidate $N=1156$. $m=34$. Digits are {1, 1, 5, 6}. All < 7. OK. Transformed digits are {4, 4, 8, 9}. Form $N'=4489$. Check if $N'$ is a square. $\sqrt{4489} = 67$. Yes. So $N=1156$ is a solution. $m=34, m'=67$. The digits of $N$ are {1, 1, 5, 6}. The digits of $N'$ are {4, 4, 8, 9}. All digits of $N$ are < 7. Check: $1+3=4$, $1+3=4$, $5+3=8$, $6+3=9$. Correct.

Consider $N=6561 = 81^2$. Digits are {6, 5, 6, 1}. All < 7. OK. Transformed digits are {9, 8, 9, 4}. Form $N'=9894$. Is $N'=9894$ a perfect square? $\sqrt{9894} \approx 99.46$. No.

Consider $N=1444$? No, $38^2=1444$. Digits {1,4,4,4} < 7. Transformed digits {4,7,7,7}. $N'=4777$. Not a square.

Consider $N=444$? Not a square.

Consider $N=6084 = 78^2$. Digit 8 disqualifies it.

Consider $N=6400 = 80^2$. Digits {6,4,0,0} < 7. OK. Transformed digits {9,7,3,3}. $N'=9733$. Not a square.

Consider $N=6561 = 81^2$. Digits {6,5,6,1} < 7. OK. Transformed digits {9,8,9,4}. $N'=9894$. Not a square.

Consider $N=100 = 10^2$. Digits {1,0,0} < 7. OK. Transformed digits {4,3,3}. $N'=433$. Not a square.

Consider $N=121 = 11^2$. Digits {1,2,1} < 7. OK. Transformed digits {4,5,4}. $N'=454$. Not a square.

Consider $N=144 = 12^2$. Digits {1,4,4} < 7. OK. Transformed digits {4,7,7}. $N'=477$. Not a square.

Consider $N=169$? No, 9.

Consider $N=225 = 15^2$. Digits {2,2,5} < 7. OK. Transformed digits {5,5,8}. $N'=558$. Not a square.

Consider $N=256 = 16^2$. Digits {2,5,6} < 7. OK. Transformed digits {5,8,9}. $N'=589$. Not a square.

Consider $N=324 = 18^2$. Digit 8 disqualifies it.

Consider $N=361 = 19^2$. Digit 9 disqualifies it.

Consider $N=400 = 20^2$. Digits {4,0,0} < 7. OK. Transformed digits {7,3,3}. $N'=733$. Not a square.

Consider $N=441 = 21^2$. Digits {4,4,1} < 7. OK. Transformed digits {7,7,4}. $N'=774$. Not a square.

Consider $N=529$? No, 9.

Consider $N=576$? No, 7.

Consider $N=625 = 25^2$. Digits {6,2,5} < 7. OK. Transformed digits {9,5,8}. $N'=958$. Not a square.

Consider $N=676 = 26^2$. Digits {6,7,6}. Digit 7 disqualifies it.

Consider $N=1024 = 32^2$. Digits {1,0,2,4} < 7. OK. Transformed {4,3,5,7}. $N'=4357$. Not a square.

Consider $N=1156 = 34^2$. Digits {1,1,5,6} < 7. OK. Transformed {4,4,8,9}. $N'=4489$. Yes, $67^2$. Solution!

Consider $N=1225 = 35^2$. Digits {1,2,2,5} < 7. OK. Transformed {4,5,5,8}. $N'=4558$. Not a square.

Consider $N=1369$? No, 9.

Consider $N=1444 = 38^2$. Digits {1,4,4,4} < 7. OK. Transformed {4,7,7,7}. $N'=4777$. Not a square.

Consider $N=1521 = 39^2$. No, 9.

Consider $N=1600 = 40^2$. Digits {1,6,0,0} < 7. OK. Transformed {4,9,3,3}. $N'=4933$. Not a square.

Consider $N=2025 = 45^2$. Digits {2,0,2,5} < 7. OK. Transformed {5,3,5,8}. $N'=5358$. Not a square.

Consider $N=2116 = 46^2$. Digits {2,1,1,6} < 7. OK. Transformed {5,4,4,9}. $N'=5449$. Yes, $73.8...$. No.

Consider $N=2304 = 48^2$. Digits {2,3,0,4} < 7. OK. Transformed {5,6,3,7}. $N'=5637$. Not a square.

Consider $N=2401 = 49^2$. No, 9.

Consider $N=2500 = 50^2$. Digits {2,5,0,0} < 7. OK. Transformed {5,8,3,3}. $N'=5833$. Not a square.

Consider $N=2601 = 51^2$. Digits {2,6,0,1} < 7. OK. Transformed {5,9,3,4}. $N'=5934$. Not a square.

Consider $N=3136 = 56^2$. Digits {3,1,3,6} < 7. OK. Transformed {6,4,6,9}. $N'=6469$. Not a square.

Consider $N=3364 = 58^2$. Digits {3,3,6,4} < 7. OK. Transformed {6,6,9,7}. $N'=6697$. Not a square.

Consider $N=3600 = 60^2$. Digits {3,6,0,0} < 7. OK. Transformed {6,9,3,3}. $N'=6933$. Not a square.

Consider $N=4225 = 65^2$. Digits {4,2,2,5} < 7. OK. Transformed {7,5,5,8}. $N'=7558$. Not a square.

Consider $N=4356 = 66^2$. Digits {4,3,5,6} < 7. OK. Transformed {7,6,8,9}. $N'=7689$. Not a square.

Consider $N=4624 = 68^2$. Digits {4,6,2,4} < 7. OK. Transformed {7,9,5,7}. $N'=7957$. Not a square.

Consider $N=5041 = 71^2$. Digit 7 disqualifies it.

Consider $N=5625 = 75^2$. Digit 7 disqualifies it.

Consider $N=6241 = 79^2$. Digit 7, 9 disqualify it.

Consider $N=6400 = 80^2$. Digits {6,4,0,0} < 7. OK. Transformed {9,7,3,3}. $N'=9733$. Not a square.

Consider $N=6561 = 81^2$. Digits {6,5,6,1} < 7. OK. Transformed {9,8,9,4}. $N'=9894$. Not a square.

Okay, so far, $N=1156$ is the only solution found. This suggests that such numbers might be rare.

The Classification: Unveiling the Pattern

To classify these numbers, we need more than just trial and error. We need to leverage mathematical properties. Let $N = m^2$ be such a perfect square. Let the digits of $N$ be $d_k d_{k-1} ext{ extellipsis } d_1 d_0$, where $0 ext{ extless} d_i ext{ extless} 7$ for all $i$. Let $N' = (m')^2$ be the number formed by digits $d_i' = d_i + 3$. So $N = sum_{i=0}^k d_i 10^i$ and $N' = sum_{i=0}^k (d_i+3) 10^i = sum_{i=0}^k d_i 10^i + sum_{i=0}^k 3 imes 10^i = N + 3 imes frac{10^{k+1}-1}{9}$.

This equation N' = N + 3 imes rac{10^{k+1}-1}{9} seems problematic because the number of digits in $N$ and $N'$ might not be the same. For example, if $N=1156$, $k=3$. N' = 1156 + 3 imes rac{10^4-1}{9} = 1156 + 3 imes rac{9999}{9} = 1156 + 3 imes 1111 = 1156 + 3333 = 4489. This works!

So, the relationship is (m')^2 = m^2 + 3 imes rac{10^{k+1}-1}{9}, where $k+1$ is the number of digits in $N$. Let R_{k+1} = rac{10^{k+1}-1}{9} be the repunit with $k+1$ digits (a string of $k+1$ ones). Then $(m')^2 = m^2 + 3R_{k+1}$.

We found $N=1156 = 34^2$. Here $k+1=4$. $R_4 = 1111$. $(m')^2 = 34^2 + 3 imes 1111 = 1156 + 3333 = 4489$. And $m'=67$, since $67^2 = 4489$. This fits the equation perfectly.

What about other possibilities for $k+1$?

Case $k+1 = 1$: $N$ is a single digit. $N$ must be a perfect square with a digit < 7. Possible $N$: 0, 1, 4.

• If $N=0$, $d_0=0$. $d_0+3=3$. $N'=3$. Not a square.
• If $N=1$, $d_0=1$. $d_0+3=4$. $N'=4$. Is 4 a square? Yes, $2^2$. So $N=1$ is a solution! Digits of $N$ are {1} (< 7). Digits of $N'$ are {4} (from $1+3$). Both are squares. $m=1, m'=2$.
• If $N=4$, $d_0=4$. $d_0+3=7$. $N'=7$. Not a square.

So, $N=1$ is another solution!

Case $k+1 = 2$: $N$ has two digits. $N=m^2$, digits < 7. $N'$ formed by $d_i+3$ is $(m')^2$. Last digit $d_0$ must be 1, 2, 3, 6. Possible $N$: 16, 25, 36. (1, 4, 9 are single digits).

• If $N=16 = 4^2$. Digits {1, 6}. Both < 7. OK. Transformed digits {4, 9}. $N'=49$. Is 49 a square? Yes, $7^2$. So $N=16$ is a solution! $m=4, m'=7$.
• If $N=25 = 5^2$. Digits {2, 5}. Both < 7. OK. Transformed digits {5, 8}. $N'=58$. Not a square.
• If $N=36 = 6^2$. Digits {3, 6}. Both < 7. OK. Transformed digits {6, 9}. $N'=69$. Not a square.

So, $N=16$ is a solution!

Case $k+1 = 3$: $N$ has three digits. $N=m^2$, digits < 7. Last digit $d_0 e 4, 5$. Possible $N$: $100, 121, 144, 160(?), 200, 225, 256, 300, 324(?), 361(?), 400, 441, 484(?), 529(?), 576(?), 625, 676(?). Let's filter using $d_0 e 4, 5$. So $d_0$ must be 1, 2, 3, 6. Possible $N$: $121 = 11^2$ (last digit 1), $144 = 12^2$ (last digit 4, excluded), $169$(no 9), $21(?), 225(? no 5), 256$(no 5), $324$(no 8), $361$(no 9), $400$(last 0, excluded), $441$(last 1), $484$(no 8), $529$(no 9), $576$(no 7), $625$(no 5), $676$(no 7).

Possible $N$ with 3 digits, all < 7, and last digit 1, 2, 3, 6: $121, 144$(no), $16(?), 21(?), 225$(no), $256$(no), $361$(no), $441, 625$(no), $676$(no).

Let's list squares with 3 digits: $10^2=100$ (last 0, excluded), $11^2=121$ (last 1, ok), $12^2=144$ (last 4, excluded), $13^2=169$ (9), $14^2=196$ (9), $15^2=225$ (last 5, excluded), $16^2=256$ (last 6, ok), $17^2=289$ (8,9), $18^2=324$ (last 4, excluded), $19^2=361$ (last 1, ok), $20^2=400$ (last 0, excluded), $21^2=441$ (last 1, ok), $22^2=484$ (last 4, excluded), $23^2=529$ (9), $24^2=576$ (last 6, ok), $25^2=625$ (last 5, excluded), $26^2=676$ (7,9).

Candidates for $N$ (3 digits, all < 7, last digit 1,2,3,6): $121, 144$(no), $256, 361$(no), $441, 576$.

• $N=121 = 11^2$. Digits {1, 2, 1}. OK. Transform {4, 5, 4}. $N'=454$. Not square.
• $N=144 = 12^2$. Digits {1, 4, 4}. OK. Transform {4, 7, 7}. $N'=477$. Not square.
• $N=256 = 16^2$. Digits {2, 5, 6}. OK. Transform {5, 8, 9}. $N'=589$. Not square.
• $N=441 = 21^2$. Digits {4, 4, 1}. OK. Transform {7, 7, 4}. $N'=774$. Not square.
• $N=576 = 24^2$. Digits {5, 7, 6}. Digit 7 disqualifies it.

No solutions for 3 digits.

Case $k+1 = 4$: We found $N=1156 = 34^2$. Digits {1, 1, 5, 6}. OK. Transform {4, 4, 8, 9}. $N'=4489 = 67^2$. Yes!

Let's re-check the equation $(m')^2 = m^2 + 3R_{k+1}$. For $N=1$, $k+1=1$, $R_1=1$. $1^2 = 1^2 + 3 imes 1 = 4 = 2^2$. Works. For $N=16$, $k+1=2$, $R_2=11$. $4^2 = 16$. $(m')^2 = 16 + 3 imes 11 = 16 + 33 = 49 = 7^2$. Works. For $N=1156$, $k+1=4$, $R_4=1111$. $34^2 = 1156$. $(m')^2 = 1156 + 3 imes 1111 = 1156 + 3333 = 4489 = 67^2$. Works.

Are there any other solutions? We need to check larger $k+1$.

Let $N$ be such a number. Consider its value modulo 9. The sum of digits of $N$ is $S(N)$. $N ext{ mod } 9 = S(N) ext{ mod } 9$. Since $N$ is a perfect square, $N ext{ mod } 9$ must be in {0, 1, 4, 7}. (Squares mod 9 are $0^2=0, 1^2=1, 2^2=4, 3^2=0, 4^2=16 ightarrow 7, 5^2=25 ightarrow 7, 6^2=36 ightarrow 0, 7^2=49 ightarrow 4, 8^2=64 ightarrow 1$).

Let $N'$ be the transformed number. The digits of $N'$ are $d_i' = d_i+3$. The sum of digits is S(N') = ext{sum}(d_i+3) = ext{sum}(d_i) + 3 imes ( ext{# of digits}) = S(N) + 3(k+1).

$N' ext{ mod } 9 = S(N') ext{ mod } 9 = (S(N) + 3(k+1)) ext{ mod } 9$. Since $N'$ is a perfect square, $N' ext{ mod } 9 ext{ must be in } \{0, 1, 4, 7\}$.

Let's analyze $N=1156$. $S(N)=13$. $N ext{ mod } 9 = 13 ext{ mod } 9 = 4$. This is a valid square mod 9. $k+1=4$. $S(N') = S(4489) = 25$. $N' ext{ mod } 9 = 25 ext{ mod } 9 = 7$. This is a valid square mod 9. Also, $S(N) + 3(k+1) = 13 + 3(4) = 13 + 12 = 25$. $25 ext{ mod } 9 = 7$. The calculation matches.

Let's test $N=16$. $S(N)=7$. $N ext{ mod } 9 = 7$. Valid. $k+1=2$. $S(N') = S(49)=13$. $N' ext{ mod } 9 = 13 ext{ mod } 9 = 4$. Valid. $S(N) + 3(k+1) = 7 + 3(2) = 7+6 = 13$. $13 ext{ mod } 9 = 4$. Matches.

Let's test $N=1$. $S(N)=1$. $N ext{ mod } 9 = 1$. Valid. $k+1=1$. $S(N') = S(4)=4$. $N' ext{ mod } 9 = 4$. Valid. $S(N) + 3(k+1) = 1 + 3(1) = 4$. $4 ext{ mod } 9 = 4$. Matches.

Consider the possibility of solutions with more digits. The repunits $R_{k+1}$ grow fast. $(m')^2 - m^2 = 3R_{k+1}$. $(m'-m)(m'+m) = 3R_{k+1}$.

For $k+1=4$, $(m'-m)(m'+m) = 3333$. $m=34, m'=67$. $m'-m=33$, $m'+m=101$. $33 imes 101 = 3333$. Checks out.

For $k+1=5$, $R_5=11111$. $3R_5 = 33333$. $(m'-m)(m'+m) = 33333$. We need $m^2 < 70000$ (max 5 digit number with digits < 7). $m < \sqrt{70000} \approx 264$. $m'^2 = m^2 + 33333$. If $m=264$, $m^2 = 69696$. Digits are {6,9,6,9,6}. Contains 9. Invalid. We need $m^2$ to have digits < 7. Let's try $m=258$, $m^2=66564$. Digits are {6,6,5,6,4}. OK. $N=66564$. Transformed digits {9,9,8,9,7}. $N'=99897$. Not a square. $m'= sqrt{66564+33333} = sqrt{99897} \approx 316$. So $m'$ would be around 316. If $m'=316$, $m'^2 = 99856$. Digits {9,9,8,5,6}. Not all < 7. In fact, $N'$ cannot have digits 8 or 7 and be a square if it's made of {3,4,5,6,7,8,9}. This seems to be the core constraint. Let's re-verify this. A number ending in 7, 8, 2, 3 cannot be a square. Our $N'$ digits are $d_i+3$.

If $d_i=4$, $d_i'=7$. So $N'$ cannot contain 4 unless it's the last digit (transformed from 1). But $d_0$ can't be 4. So if $d_i=4$ for $i>0$, $N'$ has a 7. Could $N'$ be a square? If $N'$ has a 7, it must be in a position where it doesn't disqualify the number.

If $d_i=5$, $d_i'=8$. $N'$ has an 8. Can $N'$ be a square? Yes, e.g. $81=9^2$. Here $N'$ has digit 8. But $N'$ cannot END in 8. So if $d_i=5$ for $i>0$, $N'$ has an 8. Can it be a square? Maybe.

If $d_i=0$, $d_i'=3$. $N'$ has a 3. Can $N'$ be a square? Yes, e.g. $36=6^2$. But $N'$ cannot END in 3. So if $d_i=0$ for $i>0$, $N'$ has a 3. Can it be a square?

If $d_i=1$, $d_i'=4$. $N'$ has a 4. Allowed. If $d_i=2$, $d_i'=5$. $N'$ has a 5. Allowed. If $d_i=3$, $d_i'=6$. $N'$ has a 6. Allowed. If $d_i=6$, $d_i'=9$. $N'$ has a 9. Allowed.

So the critical transformations $d_i ightarrow d_i'$ that yield digits {3, 7, 8} for $N'$ are when $d_i ightarrow 0, 4, 5$. The number $N'$ formed by these digits must be a square.

Consider $N=1156$. $N'$ has digits {4, 4, 8, 9}. $N'=4489=67^2$. The digit 8 is present, but it's not the last digit. The last digit is 9. This seems to be okay.

It appears that the found solutions are indeed the only ones.

Conclusion: The Perfect Squares Unveiled

After a thorough investigation, we've uncovered the perfect squares that satisfy both the digit constraint (all digits strictly less than 7) and the digit transformation rule (adding 3 to each digit results in another perfect square). These numbers are surprisingly few, highlighting the power of combining number theoretic properties with computational exploration.

Our analysis led us to three such perfect squares:

1. N = 1

   • Digits: {1}. All < 7. Condition 1 met.
   • Transform: $1+3=4$. $N'=4$. $4 = 2^2$, a perfect square. Condition 2 met.

2. N = 16

   • Digits: {1, 6}. All < 7. Condition 1 met.
   • Transform: $1+3=4$, $6+3=9$. $N'=49$. $49 = 7^2$, a perfect square. Condition 2 met.

3. N = 1156

   • Digits: {1, 1, 5, 6}. All < 7. Condition 1 met.
   • Transform: $1+3=4$, $1+3=4$, $5+3=8$, $6+3=9$. $N'=4489$. $4489 = 67^2$, a perfect square. Condition 2 met.

We systematically explored possible numbers of digits and used constraints like the possible last digits of squares and the properties of modular arithmetic. The equation $(m')^2 = m^2 + 3R_{k+1}$ provided a rigorous framework for verification. While larger numbers could theoretically exist, the constraints imposed by the digits and the structure of squares make them increasingly improbable. The rarity of these solutions underscores the elegance and depth found within elementary number theory problems. It's a testament to how seemingly simple rules can lead to profound mathematical discoveries!

Keep exploring, keep questioning, and who knows what other number patterns you'll uncover, guys, uncover!