# Show that if $(a,b)=1$ then $(a+b, a^2-ab+1)=1$ or 3.

Kindly help me to solve the following

If $(a,b)=1$ then we have to show that $(a+b, a^2-ab+1)=1$ or 3.

I managed to show $(a+b,a^2-ab+b^2)=1$ or 3. But got stuck in the above one. Please help me out.

Let $a=7,b=2$. Then note $\gcd(a,b)=1$. Also note that $$\gcd(7+2, 7^2-2 \times 7+1)=\gcd(9, 36)=9 \neq 3,1$$ So your claim does not hold.