Thread:

1. Quadratic Residues

Let $p$ be a prime such that $p>(n^2+n+k)^2+k$. ( $<br /> n,k \in \mathbb{Z}^ + <br />$ )

Consider the sequence: $<br /> n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k<br />$.

Prove that there's a pair $(m,m+k)$ of integers from our sequence, such that $<br /> \left( {\tfrac{m}<br /> {p}} \right)_L = \left( {\tfrac{{m + k}}<br /> {p}} \right)_L = 1<br />$

This is a lovely problem, have fun!.

PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too)

2. Originally Posted by PaulRS
Let $p$ be a prime such that $p>(n^2+n+k)^2+k$. ( $<br /> n,k \in \mathbb{Z}^ + <br />$ )

Consider the sequence: $<br /> n^2 ,n^2 + 1,...,\left( {n^2 + n + k} \right)^2 + k<br />$.

Prove that there's a pair $(m,m+k)$ of integers from our sequence, such that $<br /> \left( {\tfrac{m}<br /> {p}} \right)_L = \left( {\tfrac{{m + k}}<br /> {p}} \right)_L = 1<br />$

This is a lovely problem, have fun!.

PS: By the way, I'd posted 2 problems like in January, NCA gave a solution for 1, but the other is waiting! (haha, and it is a nice problem too)
if $\left(\frac{n^2+k}{p} \right)=1,$ choose $m=n^2$ and if $\left(\frac{(n+1)^2+k}{p} \right)=1,$ then choose $m=(n+1)^2.$ so we may assume that $\left(\frac{n^2+k}{p} \right)=\left(\frac{(n+1)^2+k}{p} \right)=-1.$

now we have this cute identity: $(n^2+k)((n+1)^2+k)=(n^2+n+k)^2+k.$ therefore: $\left(\frac{(n^2+n+k)^2+k}{p} \right)=1$ and we choose $m=(n^2+n+k)^2. \ \Box$

Remark: to put the above result in simple words: for any positive integers $n,k$ and any prime number $p > (n^2+n+k)^2+k,$ at least one of the numbers

$n^2+k, \ (n+1)^2+k, \ (n^2+n+k)^2+k$ is a quadratic residue modulo $p.$
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