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?
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$\begingroup$ 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."? $\endgroup$– HSNCommented Jan 19, 2017 at 16:02
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3$\begingroup$ See Bill Dubuque's answer to another recent question. $\endgroup$– Jyrki LahtonenCommented Jan 19, 2017 at 16:10
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$\begingroup$ @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 kapoorCommented Jan 19, 2017 at 16:16
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$\begingroup$ @Jyrki Lahtonen : thanks for that link $\endgroup$– deepa kapoorCommented Jan 19, 2017 at 16:20
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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 DubuqueCommented Jan 19, 2017 at 16:34
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3 Answers
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$n^5 + n^4 + 1 = n^5 + n^4 + n^3 – n^3 – n^2 − n + n^2 + n + 1$
$\implies$$ n^3(n^2 + n + 1) − n(n^2 + n + 1) + (n^2 + n + 1)$
=$ (n^2 + n + 1)(n^3 − n + 1)$
Hence, for $n>1$, $n^5 + n^4 + 1$ is not a prime number.
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$n^5+n^4+1=(n^3-n+1)(n^2+n+1)$
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7$\begingroup$ I think you answer would benefit from explaing why and how you arrived to this factorisation. $\endgroup$ Commented Jan 19, 2017 at 16:05
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$$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, 2017 at 16:08
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$\begingroup$ Can down-voter explain us why did you do it? $\endgroup$ Commented Feb 4, 2020 at 5:35