TITLE: What is the next perfect square of the form 14444... in decimal notation?
QUESTION [22 upvotes]: 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?

REPLY [7 votes]: 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.