The numbers that we want to find out are "x" and "y".
Therefore:
x²-y²=5100
x²=5100+y²
x=⁺₋√(5100+y²)
x=⁺₋√(100*51+y²)
y²=100z²
x=⁺₋√(100*51+100z²)
x=⁺₋√(100(51+z²))
x=⁺₋10√(51+z²)
(51+z²) must be equalt to: 8²,9²,10²,11²....;
Now, we have to try with this numbers:(64,81,100,121...), and z² must be a square number
1)
51+z²=64 ⇒ z²=64-51=13; this solution is not valid.
2)
51+z²=81 ⇒ z²=81-51=30; this solution is not valid.
3)
51+z²=100 ⇒ z²=100-51=49 this solution is valida, because z² is a suare number.
y²=100z²
y²=100(49)=4900
y=⁺₋√4900
We have two solutions:
y₁=-70 , so x₁=⁺₋√(5100+4900)=⁺₋100
y₂=70, so x₂=⁺₋√(5100+4900)=⁺₋100.
Answer: the first number can be: 100 or -100, and the second number can be -70 or 70.
To check:
(-100)²-(-70)²=10000-4900=5100
(100)²-(70)²=10000-4900=5100
(-100)²-(70)²=10000-4900=5100
(100)²-(-70)² =10000-4900=5100