Question 897834
This is the intersection of two parabolas. 
There are four solutions but only one of them is an integer solution.
{{{m+n^2=7}}}
{{{ m^2+n=11}}}
From eq. 1,
{{{m=7-n^2}}}
{{{m^2=49-14n^2+n^4}}}
Substituting,
{{{n^4-14n^2+49+n=11}}}
{{{n^4-14n^2+n+38=0}}}
{{{(n-2)(n^3+2n^2-10n-19)=0}}}
One integer solution:
{{{n-2=0}}}
{{{n=2}}}
Then,
{{{m+2^2=7}}}
{{{m+4=7}}}
{{{m=3}}}
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{{{drawing(300,300,-5,5,-5,5,grid(1),circle(3,2,0.3),graph(300,300,-5,5,-5,5,11-x^2,sqrt(7-x),-sqrt(7-x)))}}}