Question

How many positive integral ordered pair solutions will result for the equation
a
2 – b
2 = 120, provided a and b are positive integers?

Answer 1

Answer:

We have a2−b2=(a+b)(a−b)=1001. This implies both (a+b) and (a−b) must be factors of 1001 and (a−b)=1001/(a+b) and (a+b)>(a−b) as both a,b are positive. Thus, we just need the factors of 1001 which are larger than the square root of 1001 as only they satisfy the above condition. Since 1001 is not a perfect square exactly half of the factors are larger than the square root. Now, 1001=71×111×131 and hence it has (1+1)×(1+1)×(1+1)=8 factors. Thus we have 4 solutions.