Question 946223
{{{(N-2)^2+N^2=(N+2)^2+48}}}
{{{N^2-4N+4+N^2=N^2+4N+4+48}}}
{{{N^2-8N-48=0}}}
{{{(N-12)(N+4)=0}}}
Only the positive solution is needed.
{{{N-12=0}}}
{{{N=12}}}
So the integers are 10,12, and 14.