Question 551403
Let the greater number be x and the lesser number be y.
{{{x=2y+3}}}, so {{{x>2y>y>0}}}
Since {{{x>y>0}}}, of course {{{x^2>y^2>0}}}.
{{{x^2-y^2=24}}}
Substituting {{{x=2y+3}}} into {{{x^2-y^2=24}}}, we get
{{{(2y+3)^2-y^2=24}}} --> {{{4y^2+12y+9-y^2=24}}} --> {{{3y^2+12y-15=0}}} --> {{{y^2+4y-5=0}}} --> {{{(y-1)(y+5)=0}}}
The positive solution is {{{y=1}}}. (We ignore solution {{{y=-5}}} because we know y is positive).
Then {{{x=2y+3=2*1+3=2+3=5}}}
So the two integers are 1 and 5.