Algebra Question System of Equations

Question

How would one go about solving the system of five equations:

$p^2=p+q-2r+2s+t-8$,

$q^2=-p-2q-r+2s+2t-6$,

$r^2=3p+2q+r+2s+2t-31$,

$s^2=2p+q+r+2s+2t-2$,

$t^2=p+2q+3r+2s+t-8$

over the integers? I have no immediate way of answering this question, since it looks to be solved by some "trick." Inequalities may help, although it says "over the INTEGERS" and most inequalities only deal with positive reals.

EDIT: I dont see how congruence's can work, as that may limit the number of solutions, but only to a certain congruence class. For example, we may find that p=1 mod 3, say, but this will only give us an infinite number of p's to check. Unless, of course, we get a congruence contradiction, in which case there would be no solution, but the solution $(3,2,1,5,4)$ works-noted below. (sorry my computer is acting up and wont let me comment)

Answer 1

A start: I would start by trying to find combinations which simplify things and work modulo small primes. The first equation tells us that $q$ and $t$ are both even or both odd. The second says $p$ and $r$ are likewise the same parity, and the third says $p$ is odd. Deleted: an assertion based on $s^2-q^2$ which had an algebra error.

Answer 2

Not a direct answer but too big for a comment: Rewrite your question as $$ \begin{pmatrix} p-1 &-1&2&-2&-1&8\\ 1&q+2&1&-2&-2&6\\ -3&-2&r-1&-2&-2&31\\ -2&-1&-1&s-2&-2&2\\ -1&-2&-3&-2&t-1&8 \end{pmatrix} \begin{pmatrix} p\\q\\r\\s\\t\\1 \end{pmatrix} = 0 $$ I would go for the nullspace of this matrix which is not simplifying much but maybe allow for a cleaner search. For example, $s$ and $t$ columns look suspicious :).

EDIT: A small Matlab routine gave a solution as $\begin{pmatrix}p &q &r &s &t\end{pmatrix} = \begin{pmatrix}3 &2 &1 &5 &4\end{pmatrix}$

EDIT2 : I forgot to write that I have massaged the problem a bit by applying some row manipulations from the left which is the only detail that I wanted to stress but I wrote anything but that.

Answer 3

If you add all your equations you get
$p^2+q^2+r^2+s^2+t^2=6p+4q+2r+10s+8t-55$.

By completing the square you can rewrite this as
$(p-3)^2+(q-2)^2+(r-1)^2+(s-5)^2+(t-4)^2=9+4+1+25+16-55=0$.

Obviously, the only solution is $p=3$, $q=2$, $r=1$, $s=5$, $t=4$.