Phương trình nghiệm nguyên đa biến

Q

quangltm

~#: $y = 0 \implies x=0$

~#: $y>0 \implies 2 \nmid x$

$1 + x + x^2 + x^3 = 2^y \iff (x+1)(x^2 + 1)=2^y \implies x+1= 2^m, x^2 + 1 =2^{y-m}$ với $m \in \mathbb N \implies \left( x+1 \right) ^{2}-2\,x={2}^{2\,m}-2\,x \implies {2}^{2\,m}-2\,x={2}^{y-m} \\ \implies 2^{2m-1}-x = 2^{y-m-1} \implies 2 \mid x \lor 2m-1=0 \lor y-m-1=0 \implies y-m-1=0 \implies x=1$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom