<s>[INST] 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.
 [/INST] Observe that $(ab,16)=1$
If $a=16A+1,b=16B-1\implies a+b=16(A+B)$
If $a=16A+3,b=16B+5\implies a+b=16(A+B)+8$
If $a=16A+7,b=16B+9\implies a+b=16(A+B+1)$
So,  in all the cases,  $8\mid (a+b)$
As to find the pairs, as $ab\equiv -1\pmod {16}$ ,so $a(-b)\equiv 1$ i.e.,$-b$ is inverse of $a\pmod {16}$.
we know by this, $\lambda(16)=4\implies a^4\equiv 1\pmod {16}\implies a^{-1}\equiv a^3$
For example, $a=3,a^3=27\equiv -5\pmod {16}\implies b\equiv 5$
But $16$ is small enough to allow mental arithmetic, we just need to multiply $m,n$ where $m\le n<16$  and $(mn,16)=1$
</s>