Question 759039


{{{p^2+q^2=85}}}, 

if {{{p-q=1}}},then {{{p=q+1}}}........plug it in {{{p^2+q^2=85}}}

{{{(q+1)^2+q^2=85}}}

{{{q^2+2q+1+q^2=85}}}

{{{2q^2+2q+1=85}}}

{{{2q^2+2q=84}}}...divide all by {{{2}}}

{{{q^2+q=42}}}

{{{q^2+q-42=0}}}......use quadratic formula


 {{{q = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

 {{{q = (-1 +- sqrt( 1^2-4*1*(-42) ))/(2*1) }}}

 {{{q = (-1 +- sqrt( 1+168 ))/2 }}}

{{{q = (-1 +- sqrt(169 ))/2 }}}

{{{q = (-1 +- 13)/2 }}}

solutions:

{{{q = (-1 + 13)/2 }}}

{{{q = 12/2 }}}

{{{highlight(q=6)}}}

or

{{{q = (-1 - 13)/2 }}}

{{{q = -14/2 }}}

{{{highlight(q=-7)}}}



now find {{{p}}}

{{{p-q=1}}}

{{{p-6=1}}}

{{{p=1+6}}}

{{{highlight(p=7)}}}

or

{{{p-(-7)=1}}}

{{{p+7=1}}}

{{{p=1-7}}}

{{{highlight(p=-6)}}}

so, there are two solutions:

1. {{{highlight(p=7)}}} and {{{highlight(q=6)}}}

2. {{{highlight(p=-7)}}} and {{{highlight(q=-6)}}}


then

{{{p^2-q^2=7^2-6^2=49-36=13}}}

or

{{{p^2-q^2=(-7)^2-(-6)^2=49-36=13}}}