there exsit postive integer $x,y$ such $p\mid(x^2+y^2+n)$

Question

For any give the postive integer $n$,and for any give prime number $p$.

show that

there exsit postive integer $x,y$ such $$p\mid(x^2+y^2+n)$$

My approach is the following:

Assmue that $n=1,p=2$,we choose$(x,y)=(1,2)$ $$2\mid6=1^2+2^2+1$$

Assmue that $n=1,p=3$, we choose $(x,y)=(1,2)$ $$3\mid6=1^2+2^2+1$$ Assume that $n=1,p=5$,we choose $(x,y)=(2,5)$ $$5\mid30=2^2+5^2+1$$ Assume that $n=2,p=2$, we choose $(x,y)=(2,2)$ $$2\mid10=2^2+2^2+2$$ Assume that $n=2,p=3$ we choose $(x,y)=(2,3)$ $$3\mid15=2^2+3^2+2$$ Assume that $n=2,p=5$,we choose $(x,y)=(3,3)$ $$5\mid20=3^2+3^2+2$$ and so on

Now I'm stuck and don't know how to proceed

See http://math.stackexchange.com/questions/891861 – Bart Michels Mar 21 '15 at 15:56 — Bart Michels, Mar 21 '15 at 15:56

score 1 · Accepted Answer · answered Mar 21 '15 at 15:54

The result is easy to prove if $p=2$, so we can assume from now on that $p$ is odd.

Modulo $p$, there are $\frac{p+1}{2}$ squares, namely the $\frac{p-1}{2}$ quadratic residues of $p$, and $0$.

So modulo $p$ there are $\frac{p+1}{2}$ distinct values of $x^2$. There are also (for fixed $n$) $\frac{p+1}{2}$ distinct values of $-y^2-n$, since there are $\frac{p+1}{2}$ distinct values of $y^2$, and hence of $-y^2$.

Since $\frac{p+1}{2}+\frac{p+1}{2}=p+1\gt p$, by the Pigeonhole Principle there exist $x$ and $y$ such that $x^2\equiv -y^2-n\pmod{p}$. This implies that there are values of $x$ and $y$ such that $x^2-(-y^2-n)\equiv 0\pmod{p}$.

Counting arguments can be nice. The downside is that this does not give useful information about how to produce such $x$ and $y$ efficiently when $p$ is large. — André Nicolas, Mar 21 '15 at 16:01

there exsit postive integer $x,y$ such $p\mid(x^2+y^2+n)$

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