Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

if $n$ is an integer , prove that $n^4 + 4 n^2 + 11$ is of the form $16 k$.

And I went something like:

$$\begin{align*} n^4 +4 n^2 +11 &= n^4 + 4 n^2 + 16 -5 \\ &= ( n^4 +4 n^2 -5) + 16 \\ &= ( n^2 +5 ) ( n^2-1) +16 \end{align*}$$

So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.

But, how can we do this? If anybody has any idea of how I can improve my solution, please share it here.

share|cite|improve this question
This assertion is true if and only if $n$ is an odd integer. That is, for even $n$ the assertion fails to hold. For example, for $n = 2$, we have $n^4+4n^2+11 = 43$, which is odd. For odd $n$, you can try $n = 2q+1$. – Sangchul Lee Aug 26 '12 at 9:27
sos440's comment is better justified by noticing that if $n$ is even, then so is $n^4+4n^2$. Hence $n^4+4n^2+11$ has to be odd (since $11$ is an odd number). – funktor Aug 26 '12 at 13:24
up vote 9 down vote accepted
  • $n=2k$:


Which is not $16k$.

  • $n=2k+1$:

$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16\\=(4k^2+4k)(4k^2+4k+6)+16\\=8\underbrace{k(k+1)} _{2k}(2k^2+2k+3)+16$$

Which is $16k$.

share|cite|improve this answer
The above is incorrect as in your last line you forgot to add the $\,16\,$ at the end: $$n^4+4n^2+11=8k(k+1)(2k^2+2k+3)+16$$ Now we just have to check that $\,k(k+1)(2k^2+2k+3)\,$ is always even, getting the multiple of 16 that we were missing. – DonAntonio Aug 26 '12 at 9:58
@DonAntonio: Edited, thanks. – Gigili Aug 26 '12 at 9:59
$(4k^2+4k+6)$ is a multiple of 2 while $(4k^2+4k)$ is a multiple of 8. So for odd $n$ this is a multiple of 16. – Henry Aug 26 '12 at 10:01
@Henry: The first is a multiple of $2$ and the second $4$. – Gigili Aug 26 '12 at 10:02
$4k^2+4k=4k(k+1)$ is a multiple of 8 whether $k$ is even or odd – Henry Aug 26 '12 at 10:03

The claim is false, for example $$n=2\Longrightarrow n^4+4n^2+11=16+16+11=43$$ which is not a multiple of 16. Check your expression.

Now, if $\,n=2k+1\,$ is odd, then the claim is true, since then $$n^4+4n^2+11=8k(k+1)(2k^2+2k+3)+16$$ and since $\,8k(k+1)=0\pmod {16}\,$ no matter what parity $\,k\,$ has, we're done.

share|cite|improve this answer
Thanks, this question is from burton . and as much as I remember it was given that n is an integer ( not mentioned whether it was even or odd) but I will check it . – shrey Aug 26 '12 at 9:40
this question was from david burton's book. and as much as I remember it was mentioned that n is an integer ONLY. But I will again check to make sure otherwise there was a misprint. and I didn't check that. – shrey Aug 26 '12 at 9:49
I checked: it is problem 11 in section 2.2, page 19 (6th edition), and it is given that n is an odd integer. – DonAntonio Aug 26 '12 at 9:54
If $x = 3 \mod 4$, it doesn't necessarily follow that $x^2 = 9 \mod 16$. Take $x=7$, for example. – TonyK Aug 26 '12 at 11:57
True, but it does in this case. Anyway, corrected. Thanks. – DonAntonio Aug 26 '12 at 12:10

If $2|k=>16|n^4$ and $4|n^2=>16|(n^4+4n^2)=>n^4+4n^2+11≡11\pmod{16}$


$n$ is odd$=2k+1$(say), $n^2=(2k+1)^2=8\cdot\frac{k(k+1)}{2}+1≡1\pmod{8}=>8|(n^2-1)$

(i)So, $n^4+4n^2+11=(n^2-1)^2+6(n^2-1)+16≡0\pmod{16}$ if $n$ is odd.

(ii)When $n$ is odd, $2|(n^2+1)$ and $8|(n^2-1)$(already proved) $=>2\cdot8|(n^2-1)\cdot(n^2+1)=>16|(n^4-1) $

(iii)When $n$ is odd, $n^2≡1\pmod{8}=1+8m$(say),

So, $n^4=(n^2)^2=(1+8m)^2=1+16m+64m^2≡1\pmod{16}$

So using (ii) or (iii), $n^4≡1\pmod{16}$ and $n^2≡1\pmod{8}=>4n^2≡4\pmod{32}$ if $n$ is odd,

So, $n^4+4n^2+11≡1+4+11\pmod{16}≡0\pmod{16}$ if $n$ is odd.

Alternatively, using Carmichael Function, $\lambda(16)=\frac{\phi(16)}{2}=4$ and $\lambda(8)=\frac{\phi(8)}{2}=2$

So, $n^4≡1\pmod{16}$ and $n^2≡1\pmod{8}=>4n^2≡4\pmod{32}$ if $(16,n)=1$ i.e., $n$ is odd,

So, $n^4+4n^2+11≡0\pmod{16}$ if $n$ is odd(like (ii)).

share|cite|improve this answer
+1 Two solutions are both ingenious! – awllower Aug 26 '12 at 10:56

If $m$ is odd, then $m^2 \equiv 1$ (mod 8), since $(2a+1)^{2} = 4(a^{2}+a) +1$ and $a^2 +a$ is always even when $a$ is an integer. If $h$ is an integer congruent to $3$ (mod 8), then $h^{2}-9 = (h-3)(h+3)$ is divisible by $16$. Now when $n$ is odd, we have $n^{2}+2 \equiv 3$ (mod $8$), so $(n^{2}+2)^{2} \equiv 9$ (mod $16$), so $n^{4} + 4n^{2} + 11 \equiv 9 +7 \equiv 0$ (mod 16).

share|cite|improve this answer
Remark: this method is essentially equivalent to completing the square (as did the OP) $$\rm mod\ 16\!:\,\ n^4\!+4n^2\!+11 \,\equiv\, (n^2\!+2)^2+7\,\equiv\, (n^2\!+2)^2\!-3^2\,\equiv\, (n^2\!-1)(n^2\!+5)$$ then proceeding as in my answer. – Bill Dubuque Aug 26 '12 at 14:45
Yes, I agree there is not much difference, though I had not seen your answer when I wrote mine. – Geoff Robinson Aug 26 '12 at 14:46

Hint $\rm\: n\,$ odd $\rm\,\Rightarrow 2\:|\:n^2\!+\!5,\ 8\:|\:n^2\!-\!1\!\:\Rightarrow\:16\:|\:(n^2\!+\!5)(n^2\!-\!1)\ $ by $\rm\:mod\ 8\!:\ odd^2 \equiv \{\pm1,\pm3\}^2\equiv 1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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