Find all positive integers n for which \( n^2 - 19n + 99\) is a perfect square.

MeldHunter Jun 24, 2024

#1**0 **

We can solve this problem by considering the factorization of n2−19n+99 and checking for perfect squares.

We know that a perfect square can be factored into two identical perfect squares. So we need to find two positive integers that multiply to n2−19n+99 and are themselves perfect squares.

Here's how we can proceed:

Factoring the expression: n2−19n+99=(n−1)(n−99).

Look for factors of 99: Since 99 is itself a perfect square, we only need to consider the factor 1 and itself (99) from the factored expression.

Checking for perfect squares: We see that (n−99) is already a perfect square. Now we need to check if (n−1) is a perfect square for each possibility:

If (n−1)=12=1, then n=2.

If (n−1)=92=81, then n=82. (This doesn't satisfy the condition of n being positive)

Therefore, the only positive integer value of n for which n2−19n+99 is a perfect square is n=2.

learnmgcat Jun 25, 2024