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Positive Integers with p²=q²+p+q+2018 | IOQM

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

Question
The number of pairs of positive integers pp and qq that satisfy p2=q2+p+q+2018p^2 = q^2 + p + q + 2018.
Solution
Answer: 2
Step 1: Rearrange.
(p2−q2)−(p+q)=2018  ⟹  (p+q)(p−q)−(p+q)=2018,(p^2 - q^2) - (p + q) = 2018 \implies (p + q)(p - q) - (p + q) = 2018,
 so
(p+q)(p−q−1)=2018.(p + q)(p - q - 1) = 2018.
 Step 2: Factor 2018=2×10092018 = 2 \times 1009, where 10091009 is prime. Since p+q≥2p + q \ge 2 and p+q>p−q−1p + q > p - q - 1, the larger factor is p+qp + q, giving 2018=2018×12018 = 2018 \times 1 or 1009×21009 \times 2. Step 3: Case p+q=2018p + q = 2018, p−q−1=1p - q - 1 = 1, i.e. p−q=2p - q = 2. Then p=1010p = 1010, q=1008q = 1008. Step 4: Case p+q=1009p + q = 1009, p−q−1=2p - q - 1 = 2, i.e. p−q=3p - q = 3. Then p=506p = 506, q=503q = 503. Answer: (p,q)=(1010,1008)(p, q) = (1010, 1008) and (506,503)(506, 503).
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