Question 216490
Let x be one number. Then the other number is (x+3)
Given:
{{{x^2 + (x+3)^2 = 117}}}
{{{x^2 + x^2 + 6x + 9 = 117}}}
{{{2x^2 + 6x - 108 = 0}}}
{{{2*(x^2 + 3x - 54)=0}}}
{{{2*(x+9)(x-6) = 0}}}
So x = -9 or x=6
The problem asked for natural numbers, so the smaller number is {{{x=6}}} and the other one is {{{6+3=9}}}