Algebra (Olympiad)hardFree

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

x

such that

2x + 1

is a perfect square but none of the integers

2x + 2, 2x + 3, \ldots, 3x + 3

is a perfect square.

Solution

Answer: 4

Step 1: Let

2x + 1 = p^2

with

p

odd, so

x = \dfrac{p^2 - 1}{2}

. Step 2: Since

2x + 1 = p^2

is itself a perfect square, the next perfect square is

(p + 1)^2

. For none of

2x + 2, \ldots, 3x + 3

to be a perfect square we require

3x + 2 < (p + 1)^2

. Step 3: Substitute

x = \dfrac{p^2 - 1}{2}

and simplify:

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

. Step 4: Converting back to

x

(using

p = \sqrt{2x + 1}

) gives

x^2 - 8x - 4 < 0

, hence

4 - 2\sqrt{5} < x < 4 + 2\sqrt{5}.

Since

\sqrt{5} \approx 2.236

, the positive integer values are

x \in \{1, 2, 3, 4, 5, 6, 7, 8\}

. Step 5: Among these,

2x + 1

is a perfect square only for

x = 4

(where

2x + 1 = 9

). Step 6: Verify

x = 4

: the integers

2x + 2, \ldots, 3x + 3

are

10, 11, 12, 13, 14, 15

, none of which is a perfect square. Answer:

x = 4

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.