Find all integer solutions such that $$a+1|2a^2+9$$
Solution. I could solve this by writing $$\frac{2a^2+9}{a+1}=2a-2+\frac{11}{a+1}.$$ So, the only integer solution for the last equation are $a=10, a=-12.$
But, i want to get a solution using divisibility properties.
As is usually the case, converting the divisibility question to modular arithmetic is very helpful:
$$ 2a^2 + 9 \equiv 0 \pmod{a+1} $$
which can be simplified dramatically by using
$$ a \equiv -1 \pmod{a+1} $$
You pretty much have the right idea. Observe that $a+1 | 2(a^2-1)$. Moreover we can write $$2a^2+9=2(a^2-1)+11.$$ Therefore for $a+1$ to divide $2a^2+9$, it has to divide $11$ (because $11$ is the difference of two expression both of which are divisible by $a+1$). Thus the only values of $a+1$ are all divisors of $11$, namely $\pm 1, \pm11$. Thus $4$ solutions in all.
By elementary means, since $a+1|2a^2+2a$, it divides $2a^2+9-(2a^2+2a)=9-2a$.
Since $a+1|2a+2$, we have $a+1|(9-2a)+(2a+2)$
Hence $a+1|11$
Hence $a+1=±1$ or $a+1=±11$
Hence $a=-12,-2,0,10$.
What do you mean by using divisibility properties?
– Mathmo123 Jun 24 '14 at 16:50