Find $(a^{2^m}+1, a^{2^n}+1)$ when a is odd and a,m,n are positive integers and m is not equal to n. I know that the hcf is a multiple of two but I can't prove that it is 2 which is the answer. Plz help.
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Please see https://math.meta.stackexchange.com/questions/5020/ – Angina Seng Nov 08 '18 at 05:33
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https://math.stackexchange.com/questions/123524/fermat-numbers-are-coprime – lab bhattacharjee Nov 08 '18 at 05:51
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Since $a$ is even, both the numbers are odd, so the HCF can't be $2$. – B. Goddard Nov 08 '18 at 21:07
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Special case of here where $,2\nmid b-c,,$ so the gcd $ =(a+1,2) = 2\ $ by $a$ odd. – Bill Dubuque Nov 12 '18 at 02:55
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a is odd, let $a=2k+1$ then:
$a^{2^m}+1=(2k+1)^{2^m}+1= 2t+2=2(t+1)$
Where due to binomial theorem t is a sum with factors $(2k)^i$ and $2^m$. Similarly we have:
$a^{2^n}+1=(2k+1)^{2^n}+1= 2s+2=2(s+1)$
Where s is the sum of terms like $(2k)^i C^{2^n}_i$. Therefore we may write:
$(a^{2m}+1, a^{2^n}+1)=2$
sirous
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