# A Diophantine equation

Let $a,b$ be integers, $a>b\ge 1$, $a^2(a+1)$ be divisible by $b$, and $3a^2$ be divisible by $b$.

Let us consider the following expression: $\frac {1+3a+3a^2+a^2(a-b+1)/b} {1+\frac{3}{a}+\frac{3}{a^2}+\frac{b}{a^2(a-b+1)}}$.

This fraction is always integer if $b=1$. For $b>1$, I know only one pair $a,b$ such that the fraction is integer, namely, $a=15,b=9$ (I checked all pairs with $a<10^4$).

Can anyone prove that there are no other pairs $a,b$ such that the fraction is integer? This question is related to the algebraic graph theory.

Thanks for any comments or hints!

• First thing I'd do is replace $a+1$ with $a$, should make things look a little less cluttered. – Gerry Myerson Jun 1 '11 at 12:38
• I think that fraction could use a common denominator, too – Federico Poloni Jun 1 '11 at 14:12
• @Gerry: you wouldn't replace "diofant"? – Igor Rivin Jun 1 '11 at 14:25
• I’ve taken the liberty to fix the spelling of the title. – Emil Jeřábek Jun 1 '11 at 16:24
• Sorry, I skipped the case $a=14,b=12$. – user15523 Jun 1 '11 at 16:34

If we substitute $f= \frac{a-b+1}{ab}$ and subtract $a^3$ the question becomes; when is this an integer: $$\frac{a^3 \times (f-1) \times (a^3 + \frac{1}{f})}{(a+1)^3 + \frac{1}{f} -1}$$ Note that $b=1$ implies that $f=1$ and this quotient is zero.
Using Maple I found that $a=14$, $b=12$ is a second example which satisfies your conditions.