Giải thích các bước giải:
$x^{4}$+$2^{4n+2}$=($x^{2}$)²+($2^{2n+1}$)²
==(x²)²+2x².$2^{2n+1}$+($2^{2n+1}$)²-2x².$2^{2n+1}$=(x²+ $2^{2n+1}$)²-4x².$2^{2n}$
=(x²+$2^{2n+1}$)²-(2x.$2^{n}$)²=(x²+ $2^{2n+1}$)²-(x.$2^{n+1}$)²
=(x²+$2^{2n+1}$+x.$2^{n+1}$)(x²+$2^{2n+1}$-x.$2^{n+1}$)
Để P∈Z thì )(\(\left[ \begin{array}{l}x²+2^{2n+1}+x.2^{n+1}=1\\x²+ 2^{2n+1}-x.2^{n+1}=1\end{array} \right.\) )
lại có x,n∈N
=> x²+$2^{2n+1}$+x.$2^{n+1}$>1
=>x(x-$2^{n}$)²+ $2^{2n}$=1
<=>(x-$2^{n}$)²+ $2^{2n}$=1
<=>$\left \{ {{n=0} \atop {x=1}} \right.$ (tm)
=>p=5∈Z thỏa mãn
