Question 919894
let sides of a right triangle be {{{a}}},{{{b}}}, and {{{c}}}  where {{{a}}} and {{{b}}} are legs, and {{{c}}} is hypotenuse (longest side)

by Pythagorean theorem: {{{c^2=a^2+b^2}}} 

so,if   {{{c=4x+5}}}, {{{b=4x-5}}} and {{{a=x}}}, then

 {{{(4x+5)^2=x^2+(4x-5)^2}}}

{{{(4x)^2+40x+25=x^2+(4x)^2-40x+25}}}

{{{cross((4x)^2)+40x+25=x^2+cross((4x)^2)-40x+25}}}

{{{40x+25=x^2-40x+25}}}

{{{0=x^2-40x-40x+25-25}}}

{{{x^2-80x=0}}}

{{{x(x-80)=0}}}

solutions:

{{{x=0}}} . ...........we cannot use this solution because {{{a=x}}}

or

{{{(x-80)=0}}} => {{{x=80}}}

then:
{{{highlight(a=80)}}}
{{{b=4x-5}}} => {{{b=4*80-5}}}=>{{{b=320-5}}}=>{{{highlight(b=315)}}}
{{{c=4x+5}}}=>{{{c=4*80+5}}}=>{{{c=320+5}}}=>{{{highlight(c=325)}}}

check:

{{{c^2=a^2+b^2}}}

{{{325^2=80^2+315^2}}}

{{{105625=6400+99225}}}

{{{105625=105625}}}