# Divisibility

• Sep 8th 2010, 07:20 AM
PaulinaAnna
Divisibility
There are given certain positive integer numbers $\displaystyle m,n,d$. Prove that if numbers $\displaystyle m^2n + 1$ and $\displaystyle mn^2 + 1$ are divisible by $\displaystyle d$, then numbers $\displaystyle m^3 + 1$ and $\displaystyle n^3 + 1$ are also divisible by $\displaystyle d$.
• Sep 8th 2010, 08:05 AM
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I don't have an answer, but for forum organization purposes this question was asked a week ago by someone else.

http://www.mathhelpforum.com/math-he...ty-154933.html
• Sep 8th 2010, 10:14 AM
Opalg
Quote:

Originally Posted by PaulinaAnna
There are given certain positive integer numbers $\displaystyle m,n,d$. Prove that if numbers $\displaystyle m^2n + 1$ and $\displaystyle mn^2 + 1$ are divisible by $\displaystyle d$, then numbers $\displaystyle m^3 + 1$ and $\displaystyle n^3 + 1$ are also divisible by $\displaystyle d$.

Just play around with multiples of d and the result will fall out:

$\displaystyle m^2n+1 = pd,\quad mn^2+1 = qd,$

$\displaystyle m^2n^2 +n = npd,\quad m^2n^2+m = mqd,$

$\displaystyle n-m = (np-mq)d,$ and so $\displaystyle n = m+(np-mq)d,$

$\displaystyle m^2\bigl(m+(np-mq)d\bigr) + 1 = pd,$

$\displaystyle m^3+1 = (p-m^2np+m^3q)d.$