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I am trying to prove that $ n^4 + 4^n $ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.

This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.

I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.

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3  
Write it as a difference of squares. –  Adam Hughes Jul 1 at 19:54
    
How? It's $n^4 + 4^n$ not $n^4 - 4^n$ –  Mathmo123 Jul 1 at 19:55
    
@Mathmo123: see my answer below. –  Adam Hughes Jul 1 at 20:03

4 Answers 4

up vote 15 down vote accepted

$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...

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Ah. Very neat!! –  Mathmo123 Jul 1 at 20:05
    
Originally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. –  pidude Jul 1 at 20:12

Hint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?

Let me know if you need further hints.

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1  
I originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. –  pidude Jul 1 at 19:59
1  
Ah... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor –  Mathmo123 Jul 1 at 20:01
    
Ok, so I was able to prove that. Now I'm working numbers which are multiples of 5. –  pidude Jul 1 at 20:09

I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.

  • We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
    1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
    2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication
    3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression
    4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression
    5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2
    6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential
  • We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again
    1. Treat $n^2+2^n$ as $a$
    2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer
    3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$
    4. $2^{2m}n^2=(n2^m)^2$ associative law
    5. Treat $n2^m$ as $b$
  • It implies that both $a$ and $b$ are both positive integer
  • From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$
  • Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$
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@Mathmo123 As I observe the behavior of the multiply operation, I see..

  • $v \times v$ produce $v$
  • $v \times o$ produce $v$
  • $o \times o$ produce $o$

and for the plus operation, I see..

  • $v+v$ produce $v$
  • $v+o$ produce $o$
  • $o+o$ produce $v$

where $v$ is even number and $o$ is odd number.

I do not know that anyone notice about this properties, so I personally name them even-odd multiply properties and even-odd sum properties. So, the expression $n^4 + 4^n$, since n is odd($o$). can be evaluated follow by that properties it will be isolated into:

  1. $n^4 = n \times n \times n \times n \equiv o \times o \times o \times o$ produces $o$
  2. $4^n \equiv v \times v \times \cdots \times v$ for $n$ times produces $v$
  3. $n^4+4^n \equiv o+v$ produces $o$

hence $2|(n^4 + 4^n)$, since $n$ is odd.
May I prove by this way? Any suggestions are kindly pleased.

So, I give up with this way.

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1. is wrong.... –  lhf Jul 2 at 11:25

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