How do I prove that $161616 \dots 16$ ($100$ digits) cannot be written as a perfect square? I did:
$$
16 \cdot 101010 \dots 1 \quad (99 \text{ digits}) 
$$
It's clear that $16 = 2^4$, satisfying the condition that the prime has to have an even power. But what about $101010 \dots 1$?
Also, I tried $16 \pmod 4$ but it's congruent to $0$.
(???)
 A: Hint: Consider the $161616\ldots 16$ ($100$ digits) number (or, since the $16$ factor is a perfect square, the $101010\ldots 1$ ($99$ digits) number) in modulo $9$, using that the sum of digits has the same remainder as the value when divided by $9$, to get that it's not one of the possible congruences of $\{0,1,4,7\}$ modulo $9$ of perfect squares.
A: Hint: If $1616\dots 16$ ($100$ digits) is a perfect square, so is $101010\dots 101$ ($99$ digits), since $16$ is a perfect square. Now consider the latter number: is it $\equiv 1 \pmod 8$?
A: I don't have reputation to comment, Hence, suggesting here.

*

*if a positive integer n is a perfect square then n can not be written in form of 4k+3, for k is an integer.


*Square of an integer leaves reminder 0 or 1 upon division by 4.


*More importantly , the square of any integer is either of the form 3k or 3k+1. In this case n= 16....16 (100 digit) , leaves 2 as remainder when divided by 3.  ${n = 2 (mod 3)}$, hence it doesn't seem a perfect square.


*This means no square can be written in form of ${3k+2}$, why? proof... By Division Algorithm every integer can be written in form of ${3k, 3k+1}$ or ${3k+2}$. if ${n=3k}$ then ${n^2}$ also of this form. if ${n=3k+1}$ then ${n^2=3K+1}$, where ${K=3k^2+2k}$ But if ${n=3k+2}$ then ${n^2=3K+1}$ where ${K=3k^2+4k+1}$, we've ${n}$ in form of ${3k+2}$ ..so, it can't be perfect square !
A: Just as you can cast out $9$s, you can also cast out $99$s.
Suppose we can write $N^2 = 161616\dots 16$. Where $16$ occurs $50$ times.
Casting out $99$s, you get
$$N^2 \equiv 16 + 16 +\dots + 16 \equiv 50\cdot 16 \equiv  800 
      \equiv 8 + 00 \equiv8 \pmod {99}$$
This implies $N^2 \equiv 8 \pmod 9$, which has no solution.
\begin{array}{c|c}
 N \mod 9 & N^2 \mod 9 \\
\hline
 0    & 0 \\
 1, 8 & 1 \\
 2, 7 & 4 \\
 3, 6 & 0 \\
 4, 5 & 7 \\
\hline
\end{array}
