Suppose that $n=x^2$ is a perfect square with an even number of base-10 digits. Assume that when n is written backwards, you get another perfect square $y^2$. Prove that 121|n. (Use the mod 11 divisibility test, and the fact that -1 is a quadratic non-residue mod 11.)
I know by using the mod 11 divisibility test (where you add and subtract alternating digits) that $x^2 \equiv -y^2$ (mod 11). I believe, also, that since $x^2$ is a quadratic residue mod 11 and $-y^2$ is a quadratic non-residue mod 11, by the properties of Legendre symbols it can't be true that $gcd(x^2,11)=gcd(y^2,11)=1$, so 11 divides $x^2$ and $y^2$. But after that, I'm stuck, and I don't know how to get to the conclusion that 121|n. Help, please?
Answer
as you said, It is easy to see that if $n$ has an even number of digits then the number obtained when reversing the digits is congruent to $-n\bmod 11$.
Therefore $n$ and $-n$ are quadratic residues. Suppose $n$ is not $0\bmod 11$. Now use the fact the quadratic residues form a subgroup of the multiplicative group $\mathbb Z_{11}^*$. So $n^{-1}$ is a quadratic residue, and subsequently $n^{-1}(-n)\equiv -1$ is a quadratic residue $\bmod 11$. This is a contradiction, the quadratic residues $\bmod 11$ are $0,1,4,9,5,3$.
We conclude $n\equiv 0 \bmod 11$. So $11|n$, and since $n$ is a square $11^2|n$ (since $11$ is a prime).
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