Let $p_1$, $p_2$, $p_3$ and $p$ be prime numbers. Prove that there are whole numbers $x$ and $y$ such that $y^2 \equiv p_1x^4-p_1p_2^2p_3^2 \pmod{p}$.
Can you please help me solve this number theory problem from a Bulgarian 2008 olympiad. I'm very very new to number theory and can't solve it. Thank you in advance!
If $-p_1$ is a square mod $p$ or $p|p_2p_3$, $x=0$ works.
If $p=3$ mod $4$, then $p_1(1-p_2^2p_3^2)$ square iff $p_1(p_2^2p_3^2-1)$ not square iff $p_1((p_2p_3)^4-p_2^2p_3^2)$ not square. So either $x=1$ or $x=p_2p_3$ work.
Otherwise, $p=1$ mod $4$, and $-p_1$ and $p_1$ are squares mod $p$, so it is enough to find $x$ such that $x^4-p_2^2p_3^2$ is a square mod $p$.
If $p_2p_3$ is a square mod $p$, we are obviously done. Else, we can assume wlog $p_2$ square mod $p$, so up to factoring $p_2^2$ (a fourth power) out, we just need, given a non-square $q=p_3$ mod $p$, find a square $z$ such that $z^2-q^2$ is a square.
Factoring $q$ out, it is equivalent to show that for some non-square $u=z/q$, $u^2-1$ is a square.
Assume it is impossible: then, $u \longmapsto u^2-1$ maps non-squares into non-squares.
So if $u$ is a non square, exactly one of $u \pm 1$ is a non square, say $u+1$. Then for the same reason, exactly one of $(u+1)\pm 1$ is a square, that is, $u-1$ and $u+2$ are squares. So all the non-squares are grouped into “cells” square-non square-nonsquare-square.
Since cells do not share their non-squares and contain exactly two of them, since there are $\frac{p-1}{2}$ nonsquares, there are exactly $\frac{p-1}{4}$ cells and $\frac{p-1}{4}$ top squares in cells (since $0$ belongs to no cell, the definition of top number means something).
Note that there are also exactly $\frac{p-1}{4}$ squares of non-squares, and that they all are top squares of certain cells.
So the squares of non-squares (ie squares not being fourth powers) are exactly the top numbers of cells. So all the top non-squares in cells are of the form $u^2-1$ for some non-square $u$.
Now, let $u$ be a non-square such that for some $t$, $u=\frac{t^2-1}{1+t^2}$. Since $-1$ is a square (of, say, $i$), $u^2-1=\left(\frac{2it}{t^2+1}\right)^2$, which is impossible. So for all $t$ non squares, either $t^2+1=0$ or $\frac{t^2-1}{t^2+1}$ is a square, ie $t^2+1$ is either a nonsquare or zero.
So as a consequence, every top number of cell (except possibly $-1$) is the bottom number of another cell.
So the pattern (for reasons of cardinality) in $\{1,\ldots,p-1\}$ of squares/non-squares mod $p$ is $S\ldots SNNSNNSNNSNNS \ldots SNNS$ where the last $S$ is $-1$.
Now, assume the bottom number of the first cell is $t^2 > 1$. Then $\frac{t^2-1}{t^2+1}$ is a square, so $t^2+1$ (inside the first cell, not a square) is a square. So the pattern is actually SNNSNNS...SNNS. This gives $\frac{p-1}{2}$ nonsquares and $\frac{p-1}{4}+2$ squares (counting zero). So $p=\frac{3(p-1)+8}{4}=\frac{3p+5}{4}$ so $p=5$.
$p=5$ is left to the reader as an exercise.