Problem:
If p is a prime and both roots of x2+px−444p=0 are integers, then
Answer Choices:
A. 1<p≤11
B. 11<p≤21
C. 21<p≤31
D. 31<p≤41
E. 41<p≤51
Solution:
Applying the quadratic formula to x2+px−444p=0, we find that its discriminant, p(p+1776), must be a perfect square. This implies that p+1776, and hence 1776, must be a multiple of p. Since 1776=24⋅3⋅37, it follows that p=2,3 or 37. It is easy to see that neither 2(2+1776) nor 3(3+1776) is a perfect square; so p=37, which indeed satisfies the conditions of the problem, leading to 111 and −148 as the two roots. Since 31<37≤41, the correct answer is (D).
OR
Rewrite the equation as x2=p(444−x) and observe that p∣x2 (i.e., p is a factor of the integer x2). Since p is prime, it follows that p∣x and so x=np for some integer n. Substituting, the equation becomes n2p=444−np, from which n(n+1)p=444. Since the factorization of 444 into primes is 3⋅4⋅37, it is evident that p=37,n=3 or −4, and x=111 or −148.