GATE 2026 EE GA Q.10: solution and answer

GATE 2026 · EEGA Q.10 · 2 marks · MCQ

$P$ and $Q$ are two positive integers such that $P^{2} = Q^{2} + 13$ .

The product of the numbers $P$ and $Q$ is ______

(A)
13
(B)
26
(C)
39
(D)
42

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

$P = 7$ and $Q = 6$ , so $P Q = 42$ . The answer is option (D).

Official answer: DSource: Official answer key, GATE 2026 EE (IIT Guwahati)

Rearrange the given relation and apply the difference of squares, $a^{2} - b^{2} = (a - b) (a + b)$ : $P^{2} - Q^{2} = (P - Q) (P + Q) = 13$
$P$ and $Q$ are positive integers and $P^{2} - Q^{2} = 13 > 0$ , so $P > Q$ . Hence $P - Q$ and $P + Q$ are both positive integers, and $P - Q < P + Q$ .
13 is prime, so the only way to write it as a product of two positive integers with the first smaller than the second is $1 \times 13$ . Therefore $P - Q = 1$ and $P + Q = 13$ .
Add the two equations: $2 P = 14 \Rightarrow P = 7$ . Subtract the first from the second: $2 Q = 12 \Rightarrow Q = 6$ .
Check: $7^{2} = 49$ and $6^{2} + 13 = 36 + 13 = 49$ . The pair satisfies the given condition.
Product: $P \times Q = 7 \times 6 = 42$ , which is option (D).

KEY IDEA

Factor $P^{2} - Q^{2} = (P - Q) (P + Q)$ . Since 13 is prime, the two positive integer factors can only be 1 and 13. That fixes $P - Q = 1$ and $P + Q = 13$ , so $P = 7$ and $Q = 6$ are the only solution.

WATCH OUT FOR

Seeing 13 in the question and picking option (A) 13 or (B) 26 without solving for $P$ and $Q$ . Another slip is to set $P - Q = 13$ and $P + Q = 1$ , which makes $Q$ negative. $P + Q$ must be the larger factor.

From the GATE 2026 EE paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2026 EE GA Q.10: solution and answer

Answer and solution

More Numerical questions