Factorizing

[nafistiham]

how to factorize this without using trial and error divisional method?
\[2^8+3^8\]

[Phlembac Adib Hasan]

I first tried with Fermat's little theorem but did not succeed. I needed to use a theorem of Dirichlet. Notice that \[8=(17-1)/2\]Since there is no solution of the modular equation $x^2\equiv 3 (\bmod17),$ we can say from Dirichlet's theorem that $ 3^8\equiv -1 (\bmod17).$ But $x^2\equiv 2 (mod 17)$ has solution; $x=6,11$ can do it.Again from Dirichlet's theorem, $ 2^8\equiv 1 (\bmod17).$Thus $2^8+3^8$ is divisible by $17$. The another factor can be found in the following way: write $17$ as $2^3+3^2.$ After some simple algebra we get the answer.\[ (2^3+3^2) (2^5-2^4\times 3^2-2^3\times 3^3+3^6)=17\times 401\].

[nafistiham]

thanks a lot
didn't know the beautiful Dirichlet theorem

[Phlembac Adib Hasan]

Dirichlet's theorem(the one I used here) states that if $p$ is an odd prime, $a$ is a co-prime of it and let $b=(p-1)/2$, then $a^b\equiv 1(mod p)$ happens if $x^2\equiv a(mod p)$ has solution.If this modular equation doesn't have any solution, then $a^b\equiv -1(mod p)$.Proof of this theorem can be found in Number Theory of Telung.

[sakibtanvir]

That's not a very hard thing.Dirichlet's theorem states that if $p$ is an odd prime, $a$ is a co-prime of it and let $b=(p-1)/2$, then $a^b\equiv 1(mod p)$ happens if $x^2\equiv a(mod p)$ has solution.If this modular equation doesn't have any solution, then $a^b\equiv -1(mod p)$.Proof of this theorem can be found in Number Theory of Telung.

[Phlembac Adib Hasan]

So what?It's not an iff condition.