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I am stuck with the following question given to us in an assignment:

Let $p>5$ be prime.Let $S=\{1,2,...,p-1\}$

Show that at least one of $2,5,10$ is a quadratic residue mod $p$.

Hence show that there exists two consecutive integers in $S$,both of which are quadratic residues mod $p$.

Use the previous result to show that there exists two consecutive integers in $S$,both of which are quadratic non-residues mod $p$.

I have done the proof of the first one.$(\frac{2}{p})(\frac{5}{p})=(\frac{10}{p})$ and all of them cannot be $-1$ as we will get $1=-1$ which is a contradiction.

But,I am not sure how to prove the next two results.I have tried in the following way:

For all $a\in S$,$(\frac{a}{p})(\frac{a+1}{p})=(\frac{a(a+1)}{p})$.Let us suppose that there does not exist two consecutive integers in $S$ such that both of them are quadratic residues mod $p$.Then LHS is $-1$.But then I am stuck and I could not prove the third part also.Can someone show me a way out?

Kishalay Sarkar
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  • For the second part, think about one, four, and nine. – Gerry Myerson Nov 14 '21 at 04:41
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    For the third part, think about how many quadratic residues/non-residues there are (and what must happen for nonresidues, if there are consecutive residues) – Pig Nov 14 '21 at 04:42

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