Answer :
35 has 4 divisors, hence two factor pairs: 1*35 and 5*7. Each corresponds to a set of perfect squares that differ by 35
One pair is ((35±1)/2)^2 = {17^2, 18^2} = {289, 324}
The other is ((7±5)/2)^2 = {1^2, 6^2} = {1, 36}
One pair is ((35±1)/2)^2 = {17^2, 18^2} = {289, 324}
The other is ((7±5)/2)^2 = {1^2, 6^2} = {1, 36}