Question 897396
Let {{{N-2}}},{{{N}}}, and {{{N+2}}} be the integers.
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{{{(N+2)^2=N^2+(N-2)^2-65}}}
{{{N^2+4N+4=N^2+N^2-4N+4-65}}}
{{{N^2-8N-65=0}}}
{{{(N-13)(N+5)=0}}}
Two solutions:
{{{N-13=0}}}
{{{N=13}}}
The integers are 11,13,and 15.
The other solution is already given.