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# plaz help i kno i already asked question but no ine helping me :C

0
65
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Compute \(\gcd(83^9+1,83^9+83^2+1)\)

i kno math iz 2 ez this mth not 2 ez :c plaz help

Mar 9, 2021

#1
+75
+4

\$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

Mar 9, 2021
#2
+2

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, \(gcd(83^9 + 1, 83^9 + 83^2 + 1) = \boxed{1} \).

Mar 9, 2021