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        riddles >> medium >> Divide The Expressions, Get Integers
        
(Message started by: K Sengupta on Jul 24^th, 2007, 8:14am)

Title: Divide The Expressions, Get Integers
Post by K Sengupta on Jul 24^th, 2007, 8:14am

Analytically determine all possible pairs of positive whole numbers (p, q) such that:

(p²q + p + q)/(pq² + q + 7) is a positive integer.

Title: Re: Divide The Expressions, Get Integers
Post by Eigenray on Jul 26^th, 2007, 3:18pm

Let k = (p²q+p+q)/(pq²+q+7). Then[hideb]qk - p = (q²-7p)/(pq²+q+7)
is an integer. Clearly it is < 1.

Case (I): qk-p http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif -1. That is,
7p-q² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif pq²+q+7,
so in particular 7p > pq², which means q is either 1 or 2.

If q=1, then we have
qk-p = (1-7p)/(p+8) = -7 + 57/(p+8)
is an integer, so p+8 | 57. The only divisors of 57 larger than 8 are 19 and 57, so we get the solutions (p,q) = (11,1) and (49,1).

If q=2, we similarly have 4(qk-p) = -7 + 79/(4p+9) is an integer. But 79 is prime, and 4p+9 can't be 1 or 79.

Case (II): qk-p=0. Then q²=7p, and k=p/q=q/7. This gives the infinite family of solutions (p,q) = (7k², 7k).[/hideb]