+9 Compute \(\gcd(83^9+1,83^9+83^2+1)\)
i kno math iz 2 ez this mth not 2 ez :c plaz help
+115 $83^9+1$ is 186,940,255,267,540,404.
$83^9+83^2+1$ is 186,940,255,267,547,293
The GCD is 1.
https://www.calculatorsoup.com/calculators/math/gcf-euclids-algorithm.php?input1=186940255267547293&input2=186940255267540404&action=solve
Notice that \(83^9 + 1\) and \(83^9 + 83^2 + 1\) differ by \(83^2\). Therefore, if they have a common divisor, then that divisor must also be a divisor of \(83^2\). (To see why this is true, suppose \(d\) is a divisor of \(83^9 + 1\), so that \(83^9 + 1 = dm\) for some integer \(m\); also suppose that \(d\) is a divisor of \(83^9 + 83^2 + 1\), so that \(83^9 + 83^2 + 1 = dn\) for some integer \(n\). Then \(83^2 = d(n-m)\).)
Since 83 is prime, the only (positive) divisors of \(83^1\)are 1, 83, and \(83^2\)itself. But 83 cannot be a divisor of \(83^9 + 1\) (which is clearly 1 more than a multiple of 8). Therefore, \(gcd(83^9 + 1, 83^9 + 83^2 + 1) = \boxed{1} \).
