A conjecture about quadratic residues

Question

Let $m$ be a positive integer, which is not a square of any other integer.

Then $x^2 \equiv m \pmod{m + 1}$ has no solution for any integer $x$.

It came up as an intermediate step in solution of a bigger problem. I've checked it numerically for the first few $m$, and it does seem to hold.

It's easily shown that the converse holds.

Answer 1

If $x^2 \equiv m \pmod{m+1}$, then, for some integer $n$,

\begin{align} x^2 &= m + (m+1)n \\ x^2+1 &= (m+1)(n+1) \\ \end{align}

So we need to find $m$ and $n$ so that $x^2+1 = (m+1)(n+1)$ and $m$ or $n$ is not a perfect square.

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\begin{array}{cc|ccc} x=5 & x^2+1 = 26 & m+1 &n+1 &m &n\\ \hline && 2 & 13 & \color{red}{1} & 12 \\ \hline \end{array}

$$5^2\equiv 12 \pmod{13}$$

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\begin{array}{cc|ccc} x=13 & x^2+1=170 & m+1 &n+1 &m &n\\ \hline && 2 & 85 & \color{red}{1} & 84\\ && 5 & 34 & \color{red}{4} & 33\\ && 10 & 17 & \color{red}{9} & \color{red}{16}\\ \hline \end{array}

$$13^2\equiv 84 \pmod{85}$$ $$13^2\equiv 33 \pmod{34}$$

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\begin{array}{cc|ccc} x=21 & x^2+1=442 & m+1 &n+1 &m &n\\ \hline && 2 & 221 & \color{red}{1} & 220\\ && 13 & 34 & 12 & 33\\ && 17 & 26 & \color{red}{16} & \color{red}{25}\\ \hline \end{array}

$$21^2\equiv 220 \pmod{221}$$ $$21^2\equiv 33 \pmod{34}$$ $$21^2\equiv 12 \pmod{13}$$

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