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Show that if $n>1$, then the number $z=n^4+4$ is a composite number.

I only tried to analyze the parity of the number $n$: if $n$ is even then $z$ is even; if $n$ is odd then $z$ is odd.

But the second part didn't help me much, how can I proceed?

More generally, for which values of $k$ is the number $n^k+k$ composite?

Ramen Nii-chan
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sant
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4 Answers4

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Observe that \begin{align} x^4+4=(x^2-2x+2)(x^2+2x+2). \end{align}

Jacky Chong
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For the $k=4$ case, one can simply note that

$$n^4 + 4 = (n^2 - 2n + 2)(n^2 + 2n + 2)$$

The motivation is realizing that you have something that looks like it could factor as two factors of $n^2 + ?n + 2$, since $(n^2)^2 = n^4$ and $2^2 = 4$, and trying to determine what that unknown quantity is. One observes that, letting that quantity be variables $b,c$ in two factors, that

$$(n^2 + bn + 2)(n^2 + cn + 2) = n^4 + n^3 (b+c) + n^2 (bc + 4) + n(2b+2c) + 4$$

So one requires that

$$\begin{cases} b+c = 0 \\ bc + 4 = 0 \\ 2b + 2c = 0 \end{cases}$$

though the latter equation is redundant. So you end up having two numbers whose sum is zero, and whose product is $-4$. Some trial and error readily let's one see that $b = -2$ and $c=2$. (Or I suppose you could try substitution or another method.)

That said, I'm not really sure about generalizations on the $n^k + k$ case.

  • Obviously $k=1$ is a dud. Take $n=6$. ($7$ is prime.)
  • In the case $k=2$, take $n=3$. ($11$ is prime.)
  • In the case $k=3$, take $n=2$. ($11$ is prime.)
  • In the case $k=5$, take $n=2$. ($37$ is prime.)
  • In the case $k=6$, not sure. (Nothing sticks out as one to me with small cases.)
  • In the case $k=7$, take $n=4$. ($16,391$ is prime.)
  • In the case $k=8$, take $n=3$. ($6,569$ is prime.)
  • In the case $k=9$, take $n=2$. ($521$ is prime.)
  • In the case $k=10$, not sure. (Nothing sticks out as one to me with small cases.)

So I know it's true for $k=4$, and maybe $6$ and $10$, but I'm not sure.

Not really up for trying more cases, but I'm not sure of a pattern offhand.

PrincessEev
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Hint $$z=n^4+4=n^4+4n^2+4-4n^2$$ is the difference of two squares.

N. S.
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Consider $k=27$ or $k=64$. Then $n^k+k$ is composite for all $n\ge 1$. Indeed,

$n^{27}+27= (n^9)^3+3^3= (n^9+3)\cdot(n^{18}-3n^9+9)$

$n^{64}+64= n^{64}+16n^{32}+64-16n^{32}=(n^{32}+8)^2-(4n^{16})^2=(n^{32}-4n^{16}+8)\cdot(n^{32}+4n^{16}+8)$

In general, if $k= m^m$ with $m\ge 3$, odd then $n^k+k$ is composite for all $n\ge 1$.

user84909
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