Let n, n+1: n² + (n+1)² = 365. Solve quadratic to get n = 13.

Find Two Consecutive Positive Integers Whose Sum of Squares Is 365

Question

Find two consecutive positive integers such that the sum of their squares is 365.

Solution — Step by Step

Let the two consecutive positive integers be $n$ and $n+1$ . We always choose the smaller one as $n$ — this keeps the algebra clean.

Sum of squares equals 365:

n^2 + (n+1)^2 = 365

Expand $(n+1)^2 = n^2 + 2n + 1$ :

n^2 + n^2 + 2n + 1 = 365

2n^2 + 2n + 1 = 365

Subtract 365 from both sides:

2n^2 + 2n - 364 = 0

Divide the entire equation by 2 — this is the step most students skip, making the numbers unnecessarily large:

n^2 + n - 182 = 0

We need two numbers that multiply to $-182$ and add to $+1$ . Think of factor pairs of 182: $1 \times 182$ , $2 \times 91$ , $7 \times 26$ , $13 \times 14$ .

Since $14 - 13 = 1$ , the split is $+14$ and $-13$ :

n^2 + 14n - 13n - 182 = 0

n(n + 14) - 13(n + 14) = 0

(n - 13)(n + 14) = 0

So $n = 13$ or $n = -14$ .

The question asks for positive integers, so $n = -14$ is rejected.

Therefore $n = 13$ , and the two consecutive integers are 13 and 14.

Verification: $13^2 + 14^2 = 169 + 196 = 365$ ✓

Why This Works

When we say “consecutive integers”, we’re looking at numbers that differ by exactly 1 — like 4 and 5, or 99 and 100. Representing them as $n$ and $n+1$ captures this relationship algebraically.

The quadratic we get, $n^2 + n - 182 = 0$ , will always have one positive and one negative root when the constant term is negative. The negative root is a valid mathematical solution, but it’s physically meaningless here since we need positive integers. Rejecting it isn’t arbitrary — it’s enforcing the condition stated in the problem.

Dividing by 2 early is a power move. Smaller coefficients mean smaller numbers to factorise, which reduces errors in the middle-school arithmetic where most marks slip away.

Alternative Method — Quadratic Formula

If factorisation doesn’t click immediately, use the formula directly on $n^2 + n - 182 = 0$ .

Here $a = 1$ , $b = 1$ , $c = -182$ :

n = \frac{-1 \pm \sqrt{1 + 4 \times 182}}{2} = \frac{-1 \pm \sqrt{729}}{2} = \frac{-1 \pm 27}{2}

$\sqrt{729} = 27$ is worth recognising. $27^2 = 729$ appears often in CBSE quadratic problems. Memorising perfect squares up to 30 saves roughly 45 seconds per problem in board exams.

Taking the positive root: $n = \frac{-1 + 27}{2} = \frac{26}{2} = 13$ .

Same answer, different route.

Common Mistake

Forgetting to divide by 2 before factorising. Students often try to factorise $2n^2 + 2n - 364 = 0$ directly, hunting for factors of $-728$ (that’s $2 \times -364$ ). This is harder and error-prone. Always simplify by dividing out any common factor across all three terms first — here, dividing by 2 converts the problem from “factors of 728” to “factors of 182”, which is far more manageable.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quadratic Formula

Common Mistake

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