Prove $n^5+n^4+1$ is not a prime

Question

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?

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."? — HSN, Jan 19 '17 at 16:02
@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. — deepa kapoor, Jan 19 '17 at 16:16
As Jyrki mentioned, if follows from the Lemma below, whose simple proof is in this answer
$$ x^{2}!+!x!+!1\mid x^A! +! x^B! +! x^C\ \ \ {\rm if}\ \ \ {A,B,C}\equiv {2,1,0}\pmod{!3} $$ — Bill Dubuque, Jan 19 '17 at 16:34

score 20 · Answer 1 · 2017-01-19T16:05:05.593

20

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

$\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.

edited Jan 19 '17 at 16:05

answered Jan 19 '17 at 16:02

score 7 · Answer 2 · answered Jan 19 '17 at 16:00

7

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

answered Jan 19 '17 at 16:00

apprenant

746

7

I think you answer would benefit from explaing why and how you arrived to this factorisation. – TZakrevskiy Jan 19 '17 at 16:05

score 5 · Answer 3 · answered Jan 19 '17 at 16:08

5

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

answered Jan 19 '17 at 16:08

Michael Rozenberg

194,933

Can down-voter explain us why did you do it? – Michael Rozenberg Feb 04 '20 at 05:35

Prove $n^5+n^4+1$ is not a prime

3 Answers3