Question 832270
Use the Pythagorean theorem,
{{{A^2+B^2=C^2}}}
{{{A^2+(3A-1)^2=(3A+1)^2}}}
{{{A^2+(9A^2-6A+1)=9A^2+6A+1}}}
{{{A^2-12A=0}}}
{{{A(A-12)=0}}}
A cannot equal zero so,
{{{A-12=0}}}
{{{A=12}}}
The sides are then {{{12}}}, {{{35}}},{{{ 37}}}.