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?

$\endgroup$
  • $\begingroup$ 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."? $\endgroup$ – HSN Jan 19 '17 at 16:02
  • 3
    $\begingroup$ See Bill Dubuque's answer to another recent question. $\endgroup$ – Jyrki Lahtonen Jan 19 '17 at 16:10
  • $\begingroup$ @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. $\endgroup$ – deepa kapoor Jan 19 '17 at 16:16
  • $\begingroup$ @Jyrki Lahtonen : thanks for that link $\endgroup$ – deepa kapoor Jan 19 '17 at 16:20
  • 2
    $\begingroup$ As Jyrki mentioned, if follows from the Lemma below, whose simple proof is in this answer $$ x^{2}\!+\!x\!+\!1\mid x^A\! +\! x^B\! +\! x^C\ \ \ {\rm if}\ \ \ \{A,B,C\}\equiv \{2,1,0\}\pmod{\!3} $$ $\endgroup$ – Bill Dubuque Jan 19 '17 at 16:34
19
$\begingroup$

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

$\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.

$\endgroup$
  • 3
    $\begingroup$ 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. $\endgroup$ – HSN Jan 19 '17 at 16:04
  • 1
    $\begingroup$ Thanks I have made an edit , even though it is mentioned in the question that $n>1$ $\endgroup$ – naveen dankal Jan 19 '17 at 16:47
5
$\begingroup$

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

$\endgroup$
  • 4
    $\begingroup$ I think you answer would benefit from explaing why and how you arrived to this factorisation. $\endgroup$ – TZakrevskiy Jan 19 '17 at 16:05
3
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.