# How would one go about solving the system of five equations: p 2 </msup> = p +

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?
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Ullveruxqte
Not a direct answer but too big for a comment: Rewrite your question as
$\left(\begin{array}{cccccc}p-1& -1& 2& -2& -1& 8\\ 1& q+2& 1& -2& -2& 6\\ -3& -2& r-1& -2& -2& 31\\ -2& -1& -1& s-2& -2& 2\\ -1& -2& -3& -2& t-1& 8\end{array}\right)\left(\begin{array}{c}p\\ q\\ r\\ s\\ t\\ 1\end{array}\right)=0$
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 $\left(\begin{array}{ccccc}p& q& r& s& t\end{array}\right)=\left(\begin{array}{ccccc}3& 2& 1& 5& 4\end{array}\right)$
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.
###### Not exactly what you’re looking for?
June Salas
${p}^{2}+{q}^{2}+{r}^{2}+{s}^{2}+{t}^{2}=6p+4q+2r+10s+8t-55$
By completing the square you can rewrite this as
$\left(p-3{\right)}^{2}+\left(q-2{\right)}^{2}+\left(r-1{\right)}^{2}+\left(s-5{\right)}^{2}+\left(t-4{\right)}^{2}=9+4+1+25+16-55=0$
Obviously, the only solution is $p=3$, $q=2$, $r=1$, $s=5$, $t=4$.