How many pairs of natural numbers are there such that the difference of whose squares is 63?
- (a) 3 ✓ UPSC's answer
- (b) 4
- (c) 5
- (d) 2
Why the answer is (a)
• Let the two natural numbers be $x$ and $y$ with $x > y$. The condition is $x^2 - y^2 = 63$.
• Factor the equation as $(x - y)(x + y) = 63$.
• Since $x$ and $y$ are natural numbers, $x - y$ and $x + y$ must be positive integer factors of 63.
• The factor pairs of 63 are (1, 63), (3, 21), and (7, 9). Note that (9, 7) is invalid because $x+y$ must be greater than $x-y$.
• For (1, 63): $x-y=1, x+y=63 \Rightarrow 2x=64 \Rightarrow x=32, y=31$ (Valid).
• For (3, 21): $x-y=3, x+y=21 \Rightarrow 2x=24 \Rightarrow x=12, y=9$ (Valid).
• For (7, 9): $x-y=7, x+y=9 \Rightarrow 2x=16 \Rightarrow x=8, y=1$ (Valid).
• There are exactly 3 such pairs: (32, 31), (12, 9), and (8, 1).
Why the other options are wrong
- (b) 4
- There are only 3 valid factor pairs of 63 that yield integer solutions for x and y, not 4.
- (c) 5
- The number of valid factor pairs is 3, not 5, as only three pairs of factors have the same parity.
- (d) 2
- There are 3 valid pairs, not 2, as the factor pair (7, 9) also yields a valid solution (8, 1).
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2020, held on 4 October 2020. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.