InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Suppose that m and n are integers such that both the quadratic equations x2 + mx - n = 0 and x2 - mx + n = 0 have integer roots. Prove that n is divisible by 6. |
|
Answer» Let a be an integer. If a is not divisible by 3 then a2 ≡ 1 (mod 3), i.e., 3 divides a2 - 1, and if a is odd then a2 ≡ 1 (mod 8), i.e., 8 divides a2 - 1. Note that the discriminants of the two quadratic polynomials are both squares of integers. Let a and b be integers such that m2 - 4n = a2 and m2 + 4n = b2. Therefore 8n = b2 - a2 and 2m2 = a2 + b2. If 3divides m then 3divides both a and b, so 3divides n. On the other hand if 3does not divide m then 3does not divide a or b. Therefore 3divides b2 - a2 and hence 3divides n. If m is odd, then so is a, and therefore 4n = m2 - a2 is divisible by 8, so n is even. On the other hand, if m is even then both a and b are even. Further (m=2)2 - n = (a=2)2 and (m=2)2 + n = (b=2)2, so (b - a) = 2 is even. In particular, n = (b2 - a2) = 4 is even. |
|