A problem about multiples. 
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a  multiple of $p$. Find the largest possible value of $p$.

I have no idea how to solve this. Please help. Thank you.

Let $ab\equiv-1\pmod {2^k}$
If $a=1,b\equiv-1= c2^k-1,a+b=c2^k$, the minimum value of $c$ is $1$.
So, $p$ must divide $2^k$, so can not have any odd factor.
Now, $b\equiv -\frac 1 a\pmod {2^k}$ as $ab\equiv-1$
So, $b\equiv -\frac 1 a\pmod {2^t}$
Let $2^t||(a+b)$ clearly, $t\le k$ 
So,$a\equiv -b \pmod{2^t}\equiv \frac 1 a \implies a^2\equiv 1\pmod{2^t}$ as $(a,2)=1$
As $a$ is odd, $=2d+1$ (say), $a^2-1=8\frac{d(d+1)}2\implies 8\mid (a^2-1) ∀$ odd $a$.
Now $\frac{d(d+1)}2$ will be even if $4\mid d(d+1)\implies d=4e$ or $d+1=4e$ as $(d,d+1)=1$
So, $\frac{d(d+1)}2$ will be odd if $d=4e+1$ or if $d=4e+2$
If $d=4e+1,a=2d+1=8e+3,a^2-1=64e^2+48e+8=8(8e^2+6e+1)$
If $d=4e+2,a=2d+1=8e+5,a^2-1=64e^2+80e+24=8(8e^2+10e+3)$
In a more compact way,if $a=8e\pm 3, a^2-1=(8e\pm 3)^2-1=64e^2\pm48c+8=8(8c^2\pm6c+1)$
$\implies 2^3||(a^2-1)$ if $a\equiv  3,5\pmod 8$
$\implies p=8  ∀ k\ge 3$  as $k=2\implies a+b=4c$
