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   Find The Pairs (X, Y)
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   Author  Topic: Find The Pairs (X, Y)  (Read 1804 times)
K Sengupta
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Find The Pairs (X, Y)  
« on: Aug 2nd, 2007, 8:01am »
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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
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Re: Find The Pairs (X, Y)  
« Reply #1 on: Jun 21st, 2008, 4:41pm »
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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.
 
 Smiley
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