8
$\begingroup$

I have to prove that for any $n>1$, the number $n^5+n^4+1$ is not a prime.With induction I have been able to show that it is true for base case $n=2$, since $n>1$.However, I cannot break down the given expression involving fifth and fourth power into simpler terms. Any help?

  • 0
    What exactly do you mean by "With induction I have been able to show that it is true for base case n=2, since n>1."? – 2017-01-19
  • 3
    See [Bill Dubuque's answer](http://math.stackexchange.com/a/2012356/11619) to another recent question. – 2017-01-19
  • 0
    @HSN: since n>1, i took the base case as n=2 and found that the given expression is equal to 49 which is certainly not a prime. Hence, the proposition is true for n=2. – 2017-01-19
  • 0
    @Jyrki Lahtonen : thanks for that link – 2017-01-19
  • 2
    As Jyrki mentioned, if follows from the Lemma below, whose simple proof is in [this answer](http://math.stackexchange.com/a/2012356/242) $$ x^{2}\!+\!x\!+\!1\mid x^A\! +\! x^B\! +\! x^C\ \ \ {\rm if}\ \ \ \{A,B,C\}\equiv \{2,1,0\}\pmod{\!3} $$ – 2017-01-19

3 Answers 3

19

$n^5 + n^4 + 1 = n^5 + n^4 + n^3 – n^3 – n^2 − n + n^2 + n + 1$

$\implies$$ n^3(n^2 + n + 1) − n(n^2 + n + 1) + (n^2 + n + 1)$

=$ (n^2 + n + 1)(n^3 − n + 1)$

Hence, for $n>1$, $n^5 + n^4 + 1$ is not a prime number.

  • 3
    You might want to add that neither $n^2+n+1$, nor $n^3-n+1$ can be 1 for $n>1$. Else it could still be a prime number. – 2017-01-19
  • 1
    Thanks I have made an edit , even though it is mentioned in the question that $n>1$ – 2017-01-19
5

$n^5+n^4+1=(n^3-n+1)(n^2+n+1)$

  • 4
    I think you answer would benefit from explaing *why* and *how* you arrived to this factorisation. – 2017-01-19
3

$$n^5+n^4+1=n^5-n^2+n^4+n^2+1=n^2(n-1)(n^2+n+1)+(n^2+n+1)(n^2-n+1)=$$ $$=(n^2+n+1)(n^3-n^2+n^2-n+1)=(n^2+n+1)(n^3-n+1)$$ I think, the best way is the following: $$n^5+n^4+1=n^5+n^4+n^3-(n^3-1)=(n^2+n+1)(n^3-n+1)$$