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Algebra (Olympiad): Find Positive Integers Perfect Square None Integers Perfect

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

Question
Find all positive integers xx such that 2x+12x + 1 is a perfect square but none of the integers 2x+2,2x+3,…,3x+32x + 2, 2x + 3, \ldots, 3x + 3 is a perfect square.
Solution
Answer: 4
Step 1: Let 2x+1=p22x + 1 = p^2 with pp odd, so x=p2−12x = \dfrac{p^2 - 1}{2}. Step 2: Since 2x+1=p22x + 1 = p^2 is itself a perfect square, the next perfect square is (p+1)2(p + 1)^2. For none of 2x+2,…,3x+32x + 2, \ldots, 3x + 3 to be a perfect square we require 3x+2<(p+1)23x + 2 < (p + 1)^2. Step 3: Substitute x=p2−12x = \dfrac{p^2 - 1}{2} and simplify:
3⋅p2−12+2<p2+2p+1  ⟹  3p2+1<2p2+4p+2  ⟹  p2−4p−1<0,3 \cdot \frac{p^2 - 1}{2} + 2 < p^2 + 2p + 1 \implies 3p^2 + 1 < 2p^2 + 4p + 2 \implies p^2 - 4p - 1 < 0,
 that is (p−2)2<5(p - 2)^2 < 5. Step 4: Converting back to xx (using p=2x+1p = \sqrt{2x + 1}) gives x2−8x−4<0x^2 - 8x - 4 < 0, hence
4−25<x<4+25.4 - 2\sqrt{5} < x < 4 + 2\sqrt{5}.
 Since 5≈2.236\sqrt{5} \approx 2.236, the positive integer values are x∈{1,2,3,4,5,6,7,8}x \in \{1, 2, 3, 4, 5, 6, 7, 8\}. Step 5: Among these, 2x+12x + 1 is a perfect square only for x=4x = 4 (where 2x+1=92x + 1 = 9). Step 6: Verify x=4x = 4: the integers 2x+2,…,3x+32x + 2, \ldots, 3x + 3 are 10,11,12,13,14,1510, 11, 12, 13, 14, 15, none of which is a perfect square. Answer: x=4x = 4.
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