Question

[Julia Nerita]

tìm số tự nhiên x y sao cho (x^2+1)(x+1)=4^y

Answer 1

[HD Film]

+) \(\left(x^2+1\right)\left(x+1\right)=4^y\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=2^{2y}\)

+) Do  \(x,y\inℕ\)nên ta có \(x^2+1=2^m\)và \(x+1=2^n\)với \(m+n=2y;m,n\inℕ\)

+) Lúc đó ta có: \(\orbr{\begin{cases}x^2+1⋮x+1\\x+1⋮x^2+1\end{cases}}\)

TH1: \(x^2+1⋮x+1\Leftrightarrow\left(x+1\right)^2-2\left(x+1\right)+2⋮x+1\)

\(\Leftrightarrow2⋮x+1\Leftrightarrow x\in\left\{0;1\right\}\)

TH2: \(x+1⋮x^2+1\Leftrightarrow x^2-1⋮x^2+1\Leftrightarrow2⋮x+1\)

\(\Leftrightarrow x\in\left\{0;1\right\}\)

* Nếu x = 0 thì \(4^y=1\Leftrightarrow y=0\)

* Nếu y = 0 thì \(4^y=4\Leftrightarrow y=1\)

Vậy \(\left(x;y\right)\in\left\{\left(0;0\right);\left(1;1\right)\right\}\)