Question 1162614
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Integers could be as  n-1, n, n+1.


{{{((n-1)+n)^2=105+(n+1)^2}}}
-
{{{(2n-1)^2=105+n^2+2n+1}}}

{{{4n^2-4n+1=105+n^2+2n+1}}}

{{{3n^2-4n-2n=105}}}

{{{3n^2-6n=105}}}

{{{n^2-2n=35}}}

{{{n(n-2)=35}}}


This equation has TWO SOLUTIONS:  n = 7  and  n = -5.


So, there are TWO OPPORTUNITIES for the triple of the numbers:


    one triple is         6,  7,  8.

    the other triple is  -6, -5, -4.
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