Giải pt nghiệm nguyên dương $9^m-3^m=n^4+2n^3+n^2+2n$
#1
Đã gửi 21-02-2013 - 12:36
- DarkBlood yêu thích
Discovery is a child’s privilege. I mean the small child, the child who is not afraid to be wrong, to look silly, to not be serious, and to act differently from everyone else. He is also not afraid that the things he is interested in are in bad taste or turn out to be different from his expectations, from what they should be, or rather he is not afraid of what they actually are. He ignores the silent and flawless consensus that is part of the air we breathe – the consensus of all the people who are, or are reputed to be, reasonable.
Grothendieck, Récoltes et Semailles (“Crops and Seeds”).
#2
Đã gửi 09-03-2013 - 21:02
Chuyên Vĩnh Phúc
#3
Đã gửi 09-03-2013 - 21:54
Ta có $$9^m-3^m=n^4+2n^3+n^2+2n$$PP. Giải phương trình nghiệm nguyên dương $$9^m-3^m=n^4+2n^3+n^2+2n$$
$$\Leftrightarrow 3^m(3^m-1)=n(n+2)(n^2+1)$$
Mà $n;n+2;n^2+1$ không có cùng chung ước là 3 nên chỉ có 1 cái chia hết cho $3^m$
Dễ thấy $3^m(3^m-1)=(n^2+2n)(n^2+1)$ có $(n^2+2n)-(n^2+1)=2n-1=3^m-(3^m-1)=1$
$\Rightarrow n=1 \Rightarrow m=1$ thử lại thấy thỏa mãn
- Zaraki và hoanglon7889 thích
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