$7^{6} | (a+b+ab)^2$ Find the value of $a,b$

Question

$7^{6} | (a+b+ab)^2$ Find the value of a,b.

I have used trial and error for a singular solution. But a generalized solution will be helpful.

Provide me the concept to deal with this problem and the approach in detail. So, that I can solve problems like this in future.

Thanks in advance!

Answer 1

Note the useful identity $$ab+a+b=(a+1)(b+1)-1.$$ We want this to be divisible by $7^3$. Easiest is to let $(a+1)(b+1)=344$.

Partial generalization: We want $(a+1)(b+1)\equiv 1\pmod{7^3}$. Let $a+1$ be almost arbitrary, just not divisible by $7$. Using the Euclidean Algorithm, or otherwise, find an inverse for $a+1$ modulo $7^3$, and set that equal to $b+1$.

Answer 2

$$7^{6}|(a+b+ab)^2$$

Let the number $ab+a+b:=k$. Now, since $7$ is a prime, the prime factorisation of $k$ should contain some $7^\alpha$ where $\alpha\ge 3$

So, $$7^{3}|a+b+ab\implies 7^3|(1+a)(1+b)-1\implies (1+a)(1+b)=1+7^3k$$ for non-negative integral $k$.

Now, check for all the factors of $1+7^3k$ and these will be the values $\{1+a,1+b\}$ take.

For $k=1$, you arrive at André's $(a+1)(b+1)=344$