Can it be prime? #4

{ n }^{ 4 }+4={ n }^{ 4 }+4{ n }^{ 2 }+4-4{ n }^{ 2 }={ ({ n }^{ 2 } }+2)^{ 2 }-{ (2n) }^{ 2 }=({ n }^{ 2 }-2n+2)({ n }^{ 2 }+2n+2)=p(p\quad is\quad prime)\\ If\quad \quad { n }^{ 2 }-2n+2=1,{ n }^{ 2 }+2n+2=p,\gg \quad n=1\quad and\quad p=5\quad ,so\quad this\quad is\quad a\quad solution\\ If\quad { n }^{ 2 }-2n+2=p,\quad { n }^{ 2 }+2n+2=1\gg n=-1\quad and\quad p=5,so\quad this\quad is\quad a\quad solution\\ If\quad { n }^{ 2 }-2n+2=-1\quad ,{ n }^{ 2 }+2n+2=-p\gg n\quad is\quad not\quad real\\ If\quad { n }^{ 2 }-2n+2=-p,\quad { n }^{ 2 }+2n+2=-1\gg n\quad is\quad not\quad real\\ So\quad the\quad solutions\quad are\quad n=1\quad and\quad n=-1,\quad so\quad the\quad answer\quad is\quad \boxed { 2 } .

$n^4 +4 =(n^2)^2 +2^2 =(n^2 +2)^2 -2(2)(n^2)=(n^2 +2)^2 -(2n)^2$ we know $a^2 -b^2 =(a+b)(a-b)$ $(n^2 +2)^2 -(2n)^2= (n^2 -2n+2)(n^2 +2n+2)$ notice that if either one is $\ne 1,-1$ than cant be a prime( only to integers) , so $n^2 -2n +2 =1, n=\dfrac{2 \pm \sqrt{4-4}}{2} =1$ now, $n^2 +2n+2=5$ , if $n=1$ , which is a prime, $n^2 +2n+2=1, n= \dfrac{-2 \pm \sqrt{4-4}}{2}=-1$ now $n^2 -2n+2=5$ if $n=-1$ , these are the only solutions since these are the only 2 with $1 \times prime$ , $n^2 -2n+2=-1,n^2 +2n+2=-1$ are complex, and therefore are rejected.

We can use Sophie-Germain identity $n^4+4r^4$ where $r=1$ . The rest you know how solve it!

$n \equiv 1,3,5,7,9 \pmod{10}$

$Case 1:$

$n \equiv 3,7,9 \pmod{10}$

$n^4+4>5$

$n^4 \equiv 1 \pmod{10}$

$n^4+4 \equiv 5 \pmod{10}$

$n^4+4 \equiv 0 \pmod{5}$

So,

$n \equiv 1 \pmod{10}$

$n^4+4 \equiv 5 \pmod{10}$

$n^4+4 \equiv 0 \pmod{5}$

The only prime which is divisble by 5 is 5 itself.

So $n= \boxed{1,-1}$

$\begin{aligned} n^4+4& = (n^2)^2 +2 \cdot 2n^2 - 2 \cdot 2n^2 + 2^2\\ &= (n^2+2)^2-(2n)^2\\ & =(n^2+2n+2)(n^2-2n+2) \\ &=((n+1)^2+1)((n-1)^2+1)\\ \end{aligned}$

$((n+1)^2+1)$ and $((n-1)^2+1)$ are both $\neq1$ iff $(n \pm 1)^2 \neq 0$

Then, $n= \{1,-1 \}$

That give us $n^4+4 =5$

So, our solutions are $n=1$ and $n=-1$