Author |
Topic: Find The Pairs (X, Y) (Read 1804 times) |
|
K Sengupta
Senior Riddler
Gender:
Posts: 371
|
|
Find The Pairs (X, Y)
« on: Aug 2nd, 2007, 8:01am » |
Quote Modify
|
Analytically determine all pairs of integers (X, Y) with X>=3 and Y>=3 such that there exists infinitely many positive integers Z for which (ZY + Z2 -1) divides : (ZX + Z -1)
|
« Last Edit: Aug 2nd, 2007, 8:02am by K Sengupta » |
IP Logged |
|
|
|
Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Find The Pairs (X, Y)
« Reply #1 on: Jun 21st, 2008, 4:41pm » |
Quote Modify
|
Well, this was a fun one! I'll change notation a bit. Let F(x) = xn+x2-1, G(x) = xm+x-1. Suppose F(x) | G(x) for infinitely many integers x. Write G(x) = q(x)F(x) + r(x), where deg r < deg F. Then r(x)/F(x) 0 as x , and is an integer infinitely often. It follows r 0, so F | G as polynomials. First suppose n is even. If is a root of F, then - is too, and they must both be roots of G. But this means m = 1-, and (-)m = 1+, which is a contradiction because 1- and 1+ are neither the same nor negatives of each other. Now the hard case: n odd. First: Since F' < 0 only for 0 < x < (2/n)1/(n-2), and F(0)<0, F has a unique real root , with 0 < < 1. Let h(t)=t-t2. Since n+2 = 1, and n<2, we have h() < h(2) = h(n), and therefore 2n+-1 < n+2-1 = 0 = m+-1. It follows m < 2n. Now, let (1-x)/(1-x2-xn) = 1 - x + x2 + a3x3 + a4x4 + .... Then ak = ak-2 for 1<k<n, and ak = ak-2 + ak-n for k n. It's easy to verify that: ak = (-1)k, 0 k < n; an+2i = i, 0 i (n-1)/2; an+2i+1 = -i, 0 i (n-3)/2. Finally, note that G/F must agree with (1-x)/(1-x2-xn) for the first m-1 terms. But G/F is a polynomial of degree m-n. It follows that the n-1 consecutive terms am-n+1,...,am-1 are all 0. But m<2n, and from a0 to a2n-1 the only zeros are an and an+1. Therefore m-n+1=n, m-1=n+1, giving n=3, m=5, and we check that indeed G/F = 1-x+x2.
|
|
IP Logged |
|
|
|
|