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Find n with n+20 and n-21 Both Perfect Squares | IOQM

JEE Maths question with a full step-by-step solution.

Question
If n+20n + 20 and n−21n - 21 are both perfect squares, where nn is a natural number, find nn.
Solution
Answer: 421
Step 1: Set up squares. Let n+20=p2n + 20 = p^2 and n−21=q2n - 21 = q^2, where pp and qq are positive integers. Step 2: Subtract the two relations.
p2−q2=(n+20)−(n−21)=41.p^2 - q^2 = (n + 20) - (n - 21) = 41.
 Step 3: Factor. (p−q)(p+q)=41(p - q)(p + q) = 41. Since 4141 is prime and 0<p−q<p+q0 < p - q < p + q,
p−q=1,p+q=41.p - q = 1, \qquad p + q = 41.
 Step 4: Solve to get p=21p = 21 and q=20q = 20. Step 5: Recover nn:
n=p2−20=441−20=421(check: n−21=400=202).n = p^2 - 20 = 441 - 20 = 421 \qquad (\text{check: } n - 21 = 400 = 20^2).
 Answer: n=421n = 421.
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