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.

Thanks in advance.

  • $\begingroup$ At least you should try this question out yourself first; this seems like a homework question... $\endgroup$ – VortexYT Mar 6 '17 at 7:30


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.

  • $\begingroup$ Thank you for the counterexample $\endgroup$ – Anjan3 Mar 6 '17 at 7:19
  • $\begingroup$ @Anjan3 Your welcome. $\endgroup$ – S.C.B. Mar 6 '17 at 7:27

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