Tìm nghiệm nguyên của phương trình :
$x^3 = 4y^3 + x^2y+y+13$
Trích lời giải Titarn trên AoPs . Khủng quá
Assume $y=0$. Then the equation is $x^3=13$ and hence no integer solution exists.
Assume $y=-1$. Then the equation is $x^3+x^2=8$. Clearly this has no integer solution.
Assume $y \ge 1$. Then denote $f(x)=x^3-yx^2-(4y^3+y+13)$. We have $f'(x)=3x^2-2xy=x(3x-2y)$ and hence $f$ has a local maximum at $x=0$ and a local minimum at $x=\frac{2y}{3}>0$. Since $f(0)=-(4y^3+y+13)<0$ we conclude that $f$ has exactly one real root.
Since $f(2y)=-y-13<0$ and $f(2y+1)=4(y-1)(2y+3) \ge 0$ the only possible integer root is $x=2y+1$ i.e. $y=1, x=3$ which is indeed a solution.
Assume $-13 \le y \le -2$. Then we have $f'(x)=x(3x-2y)$ and hence $f$ has a local minimum at $x=0$ and a local maximum at $x=\frac{2y}{3}<0$. Since $f(0)=-(4y^3+y+13)>0$ we conclude that $f$ has exactly one real root.
Since $f(2y)=-y-13 \le 0$ and $f(2y+1)=4(y-1)(2y+3) >0$ the only possible integer root is $x=2y$ i.e. $y=-13, x=-26$ which is indeed a solution.
Assume $y<-13$. As above, we still conclude that $f$ has exactly one real root. We now have $f(2y)=-y-13>0$ and $f(2y-1)=-8y^2+4y-14<0$. Hence no solution in this case.
Therefore, the only integer solutions for $(x,y)$ are $(3,1)$ and $(-26,-13)$