Consecutive Integer Problems Using Quadratic Equations

How do you form the equation?

Represent consecutive integers by n and n + 1. If their product is P, write n(n + 1) = P and rearrange to n² + n − P = 0. Solve the equation, then check whether the question asks for positive integers, any integers or another restriction.

Example: product 156

Let the smaller integer be n. Then n² + n − 156 = 0. The factor pair 13 and −12 has product −156 and sum 1, so (n + 13)(n − 12) = 0. Thus n = 12 or n = −13.

The pairs are (12, 13) and (−13, −12). Both are consecutive, and both products equal 156. If the question specifies positive integers, keep only 12 and 13. If it simply says integers, omitting the negative pair makes the answer incomplete.

Consecutive even integers use a different gap

Consecutive even integers differ by 2. If their product is 168, use n(n + 2) = 168. This gives n² + 2n − 168 = (n + 14)(n − 12), so the pairs are (12, 14) and (−14, −12). Both pairs contain even integers. Consecutive odd integers also differ by 2, but their parity must be checked separately.

What if the roots are not integers?

Not every product produces consecutive integers. For product 20, the roots of n² + n − 20 are 4 and −5, which work. For product 21, the discriminant is 85, so the roots are not integers: there is no consecutive-integer solution. Do not round a root to manufacture a pair.

Practice

1. Positive consecutive integers have product 72. 8 and 9. Check 8 × 9 = 72.
2. Positive consecutive even integers have product 120. 10 and 12. Check both the product and gap.
3. Any consecutive integers have product 30. (5, 6) and (−6, −5).

Always return the pair, not only the variable n. The meaning of the variable is part of the solution.

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