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We know that $12^2 = 144$ and that $38^2 = 1444$. Are there any other perfect squares in the form of $\frac{13}{9} (10^n - 1) + 1$ (i.e. $1$ followed by $n$ $4$'s), and how would we prove it?

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    $\begingroup$ Seems you forgot to add $1$? $\endgroup$
    – MonkeyKing
    Apr 12, 2015 at 6:33
  • $\begingroup$ Oh, yes, I suppose I did. $\endgroup$
    – Joe Z.
    Apr 12, 2015 at 6:35
  • $\begingroup$ $$(10m\pm2)^2=100m^2\pm40m+4$$ We need $100m^2\pm40m+4=\dfrac{13(10^n-1)}9+1$ $$900m^2\pm360m+40=10^n$$ $$90m^2\pm36m+4=13\cdot10^{n-1}$$ $\endgroup$ Apr 12, 2015 at 6:49
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    $\begingroup$ I would first ask whether it is possible for a square to end in 4444. If the answer is no, then you win. $\endgroup$ Apr 12, 2015 at 7:07
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    $\begingroup$ @GerryMyerson: One can easily check by brute force that 4444 is not a perfect square mod 10000, so no perfect square can end in 4444. The answer described by user128776 is much more elegant, though. $\endgroup$ Apr 12, 2015 at 19:03

3 Answers 3

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144...4 is divisible by 4 hence it follows that 144...4 is a perfect square when 3611...1 is also a perfect square.

36 and 361 are special cases because others can be written by following

3611....111 = 4(25$m$ + 2) + 3 where $m$ is in $Z$

Consider the proof in the following question:

Proving that none of these elements 11, 111, 1111, 11111...can be a perfect square

Therefore, 3611...1 can not be a perfect square.

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    $\begingroup$ Essentially, a square can't end in 11. $\endgroup$ Apr 12, 2015 at 7:09
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Brute force answer:

If $x^2 = 1\cdots4444$ then we can consider this mod $10000$ to see that we must have $x^2 \equiv 4444 \pmod{10000}$. To see if such $x$ exists, it is sufficient to consider $0 \le x < 10000$. The following C program terminates with no output, showing that no such $x$ exists.

#include <stdio.h>

int main(void) {
  int x;
  for (x = 0; x < 10000; x++) {
    if ((x * x) % 10000 == 4444) {
      printf("%d\n", x);
    }
  }
  return 0;
}

Note that you should run this on a system where int is at least 32 bits.

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  • $\begingroup$ Is there a solution for $x^2=444\pmod{1000}$ that you chose $10000$? $\endgroup$
    – Asaf Karagila
    Apr 12, 2015 at 22:15
  • $\begingroup$ @AsafKaragila: As mentioned in the question, $38^2=1444$. $\endgroup$ Apr 12, 2015 at 23:46
  • $\begingroup$ Shows you that my ability to read a question is so inconsistent, it could essentially prove and disprove the Riemann Hypothesis. $\endgroup$
    – Asaf Karagila
    Apr 12, 2015 at 23:48
  • $\begingroup$ "it is sufficient to consider" $\: 0\leq x\leq 5000 \:$, $\:$ since $\: x\mapsto x^2 \:$ is an even function. $\hspace{1.5 in}$ $\endgroup$
    – user57159
    Apr 13, 2015 at 0:02
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There are no more. No perfect square can end in $4444$ because this fails $\bmod 16$.

A number represented in base ten is congruent to its last four digits $\bmod 16$. So a number ending with four fours is congruent with $12\bmod 16$. But all squares are $\in\{0,1\}\bmod4$, and multiplying by $4$ to get an even square must then give a number $\in\{0,4\}\bmod16$. $\rightarrow\leftarrow$

In fact no square ends with four identical digits in base ten unless the quadruple terminal digit is $0$.

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