I have created a problem and I am not able to find the answer. What is the prime factorization of $15^8 + 16$? The thing is I can find an answer with a calculator easily, but what is the step by step approach to the problem.
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10Note that $15^8+16=15^8+15+1$ and $x^2+x+1$ is a factor of the polynomial $x^8+x+1$... – Steven Stadnicki Nov 29 '16 at 22:21
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2Any reason to imagine it has an especially nice factoring? In general, it isn't easy to factor numbers. In this case, it seems easy to notice that $17$ is a factor (as $2^8\equiv 1\pmod {17}$). But you are still left with a fairly large number to factor. – lulu Nov 29 '16 at 22:22
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2562890641 = 17×97×241×6449 (4 distinct prime factors) – qwr Nov 29 '16 at 22:51
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Hint $\ $ Put $\ x\!=\!15,\ I,J,K = 8,1,0\ $ below (proved in a recent question on $\,14^7\!+14^2\!+1)$
$$ \ x^{\large 2}\!+\!x\!+\!1 \,\mid\, x^I\! + x^J\! + x^K\quad {\rm if}\quad\{I,J,K\}\equiv \{2,1,0\}\!\!\!\pmod 3\,$$
Bill Dubuque
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\begin{align*} n^8+n+1 &= (n^2+n+1)(n^6-n^5+n^3-n^2+1) \\ 15^8+15+1 &= (15^2+15+1)(15^6-15^5+15^3-15^2+1) \\ &= 241 \times 10634401 \\ \end{align*}
Ng Chung Tak
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Maybe the downvote is because this factor has already been mentioned a half hour ago in a comment (and another way in an answer) so why duplicate that without adding anything new? – Bill Dubuque Nov 29 '16 at 23:01
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To be less clever than the other answers: The Pollard rho method works well here. I ran it with $f(x) =x^2+1$ and seed value $x_0=2$ and got the factor $97$ at the third step (I had to compute $x_6$. With $f(x) = x^2+2$, it gave the factor $4097$ at the fourth step.
B. Goddard
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