1987 AHSME Problems/Problem 23
Problem
If is a prime and both roots of are integers, then
Solution
For integer roots, we need the discriminant, which is , to be a perfect square. Now, this means that must divide , as if it did not, there would be a lone prime factor of , and so this expression could not possibly be a perfect square. Thus divides , which implies divides , so we must have , , or . It is easy to verify that neither nor make a perfect square, but does, so the answer is , which is .
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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