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Let $M$ be an integer, and let $p$ be a odd prime . Show that the set $$ \{M+1,M+2,M+3,\cdots,M+ 2\lfloor \sqrt{p} \rfloor+2\}$$ contains a quadratic non-residue to modulus $ p$ for all primes $ p$

maybe can use this well know: Smallest quadratic nonresidue is less than square root plus one An odd prime $p$, $a$ is the smallest positive integer that is a quadratic nonresidue modulo $p$,then $a<1+\sqrt{p}$.

maybe can sove this problem?

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    $\begingroup$ What is the point of placing a bounty on a solved problem? Is anything unclear / to be explained in my solution? $\endgroup$ – W-t-P Apr 3 '19 at 13:27
  • $\begingroup$ sorry, I don't think you're going to be able to read this. Could you be more specific?and I bounty this problem hope can see clear solution or other methods to solve it,But Thank you $\endgroup$ – function sug Apr 6 '19 at 14:59
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We know that there is a quadratic non-residue $n$ with $2\le n<\sqrt p+1$. Multiplying $n$ by the appropriate power of $4$, we can find a quadratic non-resdiue, say $n'$, in the range $\frac12\sqrt p<n'<2\sqrt p$.

Consider the set $$ S = \{Mn',(M+1)n',\dotsc,(M+2\lfloor\sqrt p\rfloor+2)n'\}\pmod p. $$ The "distance" in $\mathbb Z_p$ between any two consecutive elements of this set is $n'<2\lfloor\sqrt p\rfloor+2$, and there is no gap between the largest and the the smallest elements of the set since $$ (M+2\lfloor\sqrt p\rfloor+2)n' > Mn' + p. $$ It follows that every interval in $\mathbb Z_p$ of length $2\lfloor\sqrt p\rfloor+2$ contains at least one element of $S$. In particular, there is an element of $S$ contained in $\{M,M+1,\dotsc,M+2\lfloor\sqrt p\rfloor+2\}$. In other words, there are $x,y\in\{M,M+1,\ldots,M+2\lfloor\sqrt p\rfloor+2\}$ such that $y\equiv n'x\pmod p$. Since $n'$ is a non-residue, so is (exactly) one of $x$ and $y$.

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  • $\begingroup$ Hello, Multiplying n by the appropriate power of 4,we can find a quadratic non-resdiue, say $n'?$, it means there exist $\delta$,such $4^{\delta} n=n'?$,such $\frac{1}{2}\sqrt{p}<4^{\delta} n<2\sqrt{p}$.if so,How to sure there exist this $\delta?$ $\endgroup$ – function sug Apr 6 '19 at 14:57
  • $\begingroup$ @functionsug: Let $\delta$ be the largest integer with $4^\delta n<2\sqrt p$. Then $4^{\delta+1} n>2\sqrt p$; that is, $4\cdot 4^\delta n>2\sqrt p$, showing that $4^\delta n>\frac12\sqrt p$. Thus, $\frac12\sqrt p< 4^\delta n<2\sqrt p$, and we let $n'=4^\delta n$. $\endgroup$ – W-t-P Apr 6 '19 at 16:09
  • $\begingroup$ @functionsug: informally, you just keep multiplying $n$ by $4$ till the resulting product hits the interval $(\frac12\sqrt p,2\sqrt p)$. You cannot jump over this interval since the right endpoint of the interval is 4 times larger than its left endpoint. $\endgroup$ – W-t-P Apr 6 '19 at 20:21
  • $\begingroup$ I can't understand why demand $n'<2\sqrt{p}$What's the point of this restriction? Thanks, $\endgroup$ – function sug Apr 7 '19 at 14:13
  • $\begingroup$ For the argument to go through, we want to ensure that there is a non-residue $n'$ in the interval $(\frac12\sqrt p,2\sqrt p)$. To this end, we multiply $n$ by $4$ till the resulting product is less than $2\sqrt p$. All these multiples of $n$ are quadratic non-residues, and the largest of them will land in the target interval $(\frac12\sqrt p,2\sqrt p)$. Try to work out a numerical example, such as $p=5003$, $n=3$. $\endgroup$ – W-t-P Apr 7 '19 at 16:24

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