A Number Theory Problem by Muhammad Rasel Parvej

Assume, $P(a)=a^2-ab+b^2$ ; then the remainder of $P(a)$ when divided by $(a+b)$ , by Remainder Theorem , is $P(-b)=(-b)^2-(-b)b+b^2=3b^2$ ; the quotient, $Q(a)$ , is $(a-2b)$ ; with each of the coefficients is integer.

Now, $gcd(a^2-ab+b^2, a+b)$

$= gcd(3b^2,a+b)$

$=gcd(3,a+b)$ [As $a$ and $b$ are coprime, no factor of $b$ , other than $1$ , divides $(a+b)$ ]

So, the $gcd$ is either $3$ or $1$ . $3$ is possible, for instance, $a=2$ and $b=7$ , with $gcd(a,b)=1$ and $3| (a+b)=9$ .

Hence, the answer is $\boxed{3}$ .