Interesting mathematics problem solving question.Find all positive integers x and y such that 2x+1 is divisible by y and at the same time 2y+1 is divisible by x
i"ve tried different ways to solve this problem, and ive found values that work but I havent been able to find a solution that proves the only answers. any help is greatly appreciated! thanks :)
You need to use reminder theorem such that:
`2x +1 = p*y` (reminder is zero)
`2y + 1 = qx`
You should use the first relation to write y in terms of x such that:
`y = (2x+1)/p`
You should substitute `(2x+1)/p` for y in the second equation such that:
`2(2x+1)/p+ 1 = qx =gt 4x + 2 + p = pqx`
You need to isolate the terms in x to the left side such that:
`4x - pqx = -p-2`
Factoring out x yields:
`x(4-pq) = -p-2 =gt x(pq-4) = p+2`
`x = (p+2)/(pq-4)`
Substituting `(p+2)/(pq-4) ` for x in `y = (2x+1)/p` yields:
`y = (2(p+2)/(pq-4)+1)/p`
`y = (2p+4 +pq-4)/(p(pq-4))`
`y = p(2+q)/(p(pq-4)) =gt y = (2+q)/(pq-4)`
Hence, evaluating x and under given conditions yields x`= (p+2)/(pq-4)` and `y = (2+q)/(pq-4)` .
The answers are (1,1) (1,3) (3,1) (3,7) and (7,3).
You are not alone in wondering how to find the way of proving these are the only pairs. My teacher claims that the answers he has been given by the people behind this mathematics enrichment booklet to use as a guideline for marking does not properly prove that these are the only answers either. The proof is yet to be discovered!