Find all prime numbers that can be expressed as : $f(y)=y^{2015}+y+1$, where $y$ is a natural number. $y=1$ gives us 3, but how do we find others or prove that there can be no other??
2
-
$\begingroup$ shouldn't it be $y^{2015}$? You forgot you curly brackets. $\endgroup$– MrYouMathSep 16, 2015 at 11:30
-
$\begingroup$ I have edited it. $\endgroup$– Richard RiceSep 16, 2015 at 11:31
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Hint
Use the fact that $$x^{2015}+x+1 = x^{2015}-x^2+(x^2+x+1)=x^2(x^{3\times 671}-1)+(x^2+x+1) = x^2(x^{3}-1)(x^{3\times 670}+x^{3\times 669}+\dots+x^3+1)+(x^2+x+1) = (x^2+x+1)(x^2(x-1)A+1)$$ where $A = x^{3\times 670}+x^{3\times 669}+\dots+x^3+1$
-
-
$\begingroup$ Still not clear: $x^{2015}-x^2=x^2(x^{2013}-1)$. Where does the $3671$ come from? $\endgroup$ Sep 16, 2015 at 12:09
-
$\begingroup$ @gammatester Sorry for the confusion. With $.$ I mean product. $\endgroup$– sveSep 16, 2015 at 12:10
-
$\begingroup$ I wish you had told me "try to factorise it" as a hint. HAHA $\endgroup$ Sep 16, 2015 at 12:18