How many pairs of natural numbers are there such that the difference of whose squares is 63?
How many pairs of natural numbers are there such that the difference of whose squares is 63?
3
4
5
2
Solution
We need $a^2 - b^2 = (a-b)(a+b) = 63$ with $a, b$ natural numbers. Factor pairs of 63 with both factors of the same parity (both odd here): $1 \times 63$, $3 \times 21$, $7 \times 9$. Each gives a valid pair: $(a,b) = (32,31), (12,9), (8,1)$. So there are 3 pairs.