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$f(n)=7^{6n} - 6^{6n}$, where $n$ is a positive integer. find the divisors of $f(n)$ for odd and even values of $n$. Is there a general solution for the divisors.




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Note also that $x^3+y^3 = (x+y)(x^2-xy+y^2)$ – Mark Bennet Aug 15 '12 at 13:23
you are right Mark – Rajesh K Singh Aug 26 '12 at 8:48
up vote 2 down vote accepted

As you have said, you always have $7^{6n}-6^{6n}=(7^{3n})^2-(6^{3n})^2=(7^{3n}-6^{3n})(7^{3n}+6^{3n})$

As Mark Bennet has said $7^{3n}+6^{3n}=(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$

Also, $7^{3n}-6^{3n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})$

So we have $7^{6n}-6^{6n}=(7^n-6^n)(7^{2n}+7^n6^n+6^{2n})(7^n+6^n)(7^{2n}-7^n6^n+6^{2n})$

Alpha confirms this is the best we can do for $n=1$, but there are more for $2$ through $5$.

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For general $n$, this is asking too much. There is a formula for the irreducible factors of the polynomial $x^{6n} - y^{6n}$: there are exactly $d(6n)$ of them, corresponding to the cyclotomic polynomials dividing $6n$.

So generally $7^{6n}-6^{6n}$ will have at least $d(6n)-1$ factors in general (in this case $x-y = 1$ which discounts one factor), and one might be able to prove that some are relatively prime.

But it is hopeless to decide how often each of these algebraic factors is prime (for specific congruence classes of $n$ one could rule out primality). For instance, we have no idea whether there are infinitely many primes of the form $2^n-1$.

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