Show that gcd(a^3+1), (a^7+1)=(a+1) over the naturals.
Note: If some number divides a^3+1 and a^7+1, then it divides their difference, too.
The difference between a^3+1 and a^7+1 is
a^3(a^4 - 1)=a^3(a-1)(a+1)(a^2+1).
But clearly none of these factors divides either expression, except for (a+1), which divides both.
The factors very close to a or its power require no further elaboration, but for extra reassurance, assume that (a^2+1) divides a^3+1 for some a>1
Then (a^3+1)/(a^2+1) = (integer). But we know that (a^3+1)/((a^2+1)-a) = (a+1), differing from the required expression by (a^2+a)/(a^2+1), which is never an integer, a contradiction.
Hence gcd(a^3+1), (a^7+1)=(a+1).
Edited on May 9, 2025, 11:53 pm
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Posted by broll
on 2025-05-09 11:51:33 |