Number of solutions of simultaneous equation

Question

Let $u$ and $v$ be fixed positive constants. Then the system of equations

$$x^2+y^2=u^2+v^2$$ and $$x^3+y^3=u^3+v^3$$

(A) has infinitely many positive solutions.

(B) the number of solutions be atmost countable.

(C) has only two solutions.

(D) it is not solvable.

Clearly the pairs $x=u, y=v$ satisfies both the equations. Also $x=v, y=u$ satisfy both the equations. So there are two solutions we get. How can we conclude that there are exactly two solutions or not ?

Any hint.?

Answer 1

The hint.

Let $x+y=2p$ and $xy=q^2$.

Thus, $p^2\geq q^2$ and $$4p^2-2q^2=u^2+v^2$$ and $$8p^3-6pq^2=u^3+v^3,$$ which gives $$q^2=\frac{4p^2-u^2-v^2}{2}$$ and $$8p^3-3p(4p^2-u^2-v^2)=u^3+v^3$$ or $$4p^3-3(u^2+v^2)p+u^3+v^3=0$$ or $$(2p-u-v)(2p^2+(u+v)p-u^2+uv-v^2)=0.$$ Now, $2p=u+v$ give $(x,y)=(u,v)$ or $(x,y)=(v,u).$

Thus, it's enough to understand what happens for $2p^2+(u+v)p-u^2+uv-v^2=0.$

Can you end it now?

Answer 2

I think you can get some intuition for this problem by using calculus. Since the second equation is a strictly decreasing function of $x$, and the graph of the first equation is clearly a bounded circle, it is not too difficult to check that there must be a finite number of solutions.

Note that by implicit differentiation, the slope of the graph of $x^2 + y^2 = u^2 + v^2$ is $y' = -\frac{x}{y}$ and similarly the slope of the graph of $x^3 + y^3 = u^3 + v^3$ is $y' = -\frac{x^2}{y^2}$. If $u \neq v$, WLOG assume that $u < v$. Then at the solution point $(x,y) = (u,v)$, the slope of the cubic graph is less negative than the slope of the circle graph and so it is leaving the circle; thus it must have entered the circle for some $x < u$, giving us another solution. A similar argument shows that there is another solution with $x > v$. We can conclude that there are 4 solutions, unless $u = v$ in which case there are 3.

Answer 3

The identity $$(tx+sy)^2+(ty-sx)^2=(tx-sy)^2+(ty+sx)^2$$ where $t,s,x,y$ arbitrary shows that there are infinitely many solutions for the first system.

On the other hand, the equation $$x^3+y^3=Az^3$$ can have solutions or not. It is an elliptic curve than if have rational solutions, then excepting for the values of $A=1$ and $A=2$ have not torsion so there are infinitely many rational solutions. If you want to have solutions with $z=1$ and $x,y$ integers, there are many cases and because your system is without $A$ fixed, you can say there are many solutions.