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$$ (n + 4)^4 + 4 $$

If n is natural number, how to prove that above expression is not prime?

I am stuck here

$$ (n+4)^2 \cdot (n+4)^2 + 2 . 2 $$ $$ \left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right) $$

kdebugging
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  • Firstly, it's an expression, not an equation. Secondly, your factorization is flawed -- try multiplying it back to see. But the expression does factor. It's a well known trick. Hint: $x^4+4 = (x^2+2)^2 - 4x^2$. – quasi Apr 20 '17 at 02:21
  • if $n=-3$ then the above expression evaluates to $5$ which is prime. Do you mean that $(n+4)^2 . (n+4)^2 + 2 . 2 = (n^2+4^2 . 2)(n^2+4^2 . 2)$, could you prove this, can't you show it is not valid, and why do you need it? Use computer algebra: $(n^2+10n+26)(n^2+6n+10)$. – Mirko Apr 20 '17 at 02:30
  • @Mirko n is natural number. I will edit my question. – kdebugging Apr 20 '17 at 02:38
  • Try n = 2 and you have a counterexample. – Alex Boyle Apr 20 '17 at 03:00

3 Answers3

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From the context of the question, I assume $n ≥ 0$. Clearly for even $n$, $(n+4)^4+4$ is even. Hence suppose $n$ is odd, so $n = 2k+1$. Then $(n+4)^4+4 = (2k+5)^4+4 = 16 k^4 + 160 k^3 + 600 k^2 + 1000 k + 629 = (4 k^2 + 16 k + 17) (4 k^2 + 24 k + 37)$.

Clearly this is composite, as for $n≥0$, this is a product of two natural numbers greater than $1$.

Harambe
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Try to write a more general form as the difference of two squares which we know how to factor.

$\begin{array}\\ x^4 + a &=(x^2+u)^2-b^2x^2\\ &=x^4+2ux^2+u^2-b^2x^2\\ &=x^4+(2u-b^2)x^2+u^2\\ \text{and}\\ x^4 + a &=(x^2+u+bx)(x^2+u-bx)\\ \end{array} $

so $u^2=a$ and $b^2 = 2u$.

Therefore $b^4 = 4u^2 = 4a $.

Writing $c^2$ for $a$, this becomes

$\begin{array}\\ x^4 + c^2 &=(x^2+u)^2-b^2x^2\\ &=x^4+2ux^2+u^2-b^2x^2\\ &=x^4+(2u-b^2)x^2+u^2\\ \text{so}\\ u &=c\\ \text{and}\\ x^4 + c^2 &=(x^2+c+bx)(x^2+c-bx)\\ \end{array} $

so $b^2 = 2c$.

Since $b$ and $c$ are integers, $c = 2d^2$ and $b^2 = 4d^2$ so $b = 2d$.

Therefore, the expression can only be factored if it is of the form $x^4+4d^4$ when the factorization is $x^4+4d^4 =(x^2+2d^2)^2-4d^2x^2 =(x^2+2dx+2d^2)(x^2-2dx+2d^2) $.

This is the case $d=1$ so $b=2$ and the factorization is $x^4+4 =(x^2+2x+2)(x^2-2x+2) $.

marty cohen
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Hint: $(x+4)^4+4=0$ has roots $x = -4 \pm \sqrt{\pm 2i}=-4 \pm (1 \pm i)\,$, so the expression factors as: $$(x+4)^4+4=(x^2 + 6 x + 10) (x^2 + 10 x + 26)$$

dxiv
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