Question 1170475

if we have {{{n }}}as the first integer, then {{{n+1 }}}will be the second integer,{{{ n+2}}} will be the third integer

the lengths of the sides of a {{{right}}}{{{ triangle}}} are given by three consecutive integers so that
{{{n }}} and {{{n+1 }}} are legs
{{{ n+2}}} hypothenuse

{{{ (n+2)^2=n^2+(n+1)^2}}} .......solve for  {{{n }}}
{{{ n^2+4n+4=n^2+n^2+2n+1}}} 
{{{ n^2+4n+4=2n^2+2n+1}}} 
{{{ 0=2n^2-n^2-4n+2n+1-4}}}
{{{ 0=n^2-2n-3}}}
{{{ 0=n^2+n-3n-3}}}
{{{ 0=(n^2+n)-(3n+3)}}}
{{{ 0=n(n+1)-3(n+1)}}}
{{{ 0=(n-3)(n+1)}}}

solutions:
{{{n=3}}}
{{{n=-1}}}...disregard negative solution

the lengths of all three sides:
{{{3}}}, {{{4 }}}, and {{{ 5}}}