Show that solution $(0,0)$ is unique

Question

Assume the system: $$ \begin{align*} & 2x-5y-x(x^2+2y^2)^2=0 \\ & 5x+2y-3y(2x^2+y^2)^2=0 \end{align*} $$ Obviously, $(0,0)$ is a solution. The thing I find hard doing is showing that it's the only solution.

Attempt:

Suppose that there exist another solution $(\bar{x},\bar{y}) \neq (0,0)$. Then, for $x \neq0$ and $y \neq 0$: $$ 2x-5y=x(x^2+2y^2)^2 \iff 2-5\frac{y}{x}=(x^2+2y^2)^2=c(x,y)\geq 0 \tag{1} $$ $$ 5x+2y=3y(2x^2+y^2)^2 \iff 5 \frac{x}{y}+2=3(2x^2+y^2)^2=k(x,y) \geq 0\tag{2} $$ From equations $(1)$ and $(2)$, we yield: $$ 2-5\frac{5}{k-2}=c\iff (2-c)(k-2)=25 $$ But I can't seem to spot a contradiction. Is there another (more straightforward) way to prove this?

Answer 1

If $y=0$ so $x=0$ and we get $(0,0)$.

Let $y\neq0$ and $x=ty$.

Thus, $$\frac{2t-5}{5t+2}=\frac{t(t^2+2)^2}{3(2t^2+1)^2}$$ or $$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=0,$$ which is impossible for all real $t$.

Indeed, $$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$ $$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)>0.$$

Answer 2

The resultant of the two left sides with respect to $y$ is $$ -x \left( 531441\,{x}^{24}-2165130\,{x}^{20}+7884864\,{x}^{16}- 11409984\,{x}^{12}+12205512\,{x}^{8}+504218\,{x}^{4}+489375 \right) $$ and using Sturm's theorem one can show that $x=0$ is the only real root of that. There are of course complex roots.

EDIT: With the $3$ in $-3y$ replaced by $s$, the resultant becomes $$ \eqalign{ -x &\left( 6561\,{s}^{4}{x}^{24}-18954\,{s}^{4}{x}^{20}-23328\,{s}^{3}{ x}^{20}+21708\,{s}^{4}{x}^{16}+173340\,{s}^{3}{x}^{16}\right.\cr &+160704\,{s}^{2} {x}^{16}-23976\,{s}^{4}{x}^{12}-191016\,{s}^{3}{x}^{12}-396000\,{s}^{2 }{x}^{12}\cr & -248832\,s{x}^{12}+10692\,{s}^{4}{x}^{8}+24540\,{s}^{3}{x}^{8 }+597520\,{s}^{2}{x}^{8}+1517568\,s{x}^{8}-5832\,{s}^{4}{x}^{4} \cr &\left.+746496 \,{x}^{8}-14938\,{s}^{3}{x}^{4}+47328\,{s}^{2}{x}^{4}+337792\,s{x}^{4} -59392\,{x}^{4}+18125\,{s}^{3} \right)} $$ The discriminant of this with respect to $x$ is $$ 16618447044007316502648165488386203399239363151697251980300315466139733122351104000000000000000000000000000000000000 \,{s}^{83} \left( 295612416\,{s}^{6}-946111104\,{s}^{5}-7364205035\,{s }^{4}-13075015530\,{s}^{3}-7364205035\,{s}^{2}-946111104\,s+295612416 \right) ^{4} \left( 2772210825\,{s}^{9}-14150310144\,{s}^{8}+ 50430196864\,{s}^{7}+1618020643840\,{s}^{6}+5389314830336\,{s}^{5}+ 11451158167552\,{s}^{4}+18821256052736\,{s}^{3}+28395282890752\,{s}^{2 }+12309929918464\,s-11093095219200 \right) ^{8} $$ The roots of this polynomial should correspond to values at which the two curves are tangent. In particular, the factor $$295612416\,{s}^{6}-946111104\,{s}^{5}-7364205035\,{s}^{4}-13075015530 \,{s}^{3}-7364205035\,{s}^{2}-946111104\,s+295612416 $$ has a root at approximately $7.421911191$. Hmm: Maple indicates that the Galois group of this sextic polynomial is solvable, so the root should have an expression in radicals.

There's also a root at approximately $0.1347361851$ which makes the two curves tangent:

On the other hand, the root at $0.4013844674$ does not appear to produce tangent curves: presumably in this case the multiple root is complex.

Answer 3

I like pictures; If we increase the coefficient of $y$ in the second equation from $3$ to $7.421911190758908573490159799$ we get the first nontrivial intersections (with tangency) for this problem. The $x$ value at tangency is very close to $1,$ it may or may not actually be $1$ itself. The last picture, with no such coefficient, has rotational symmetry.

? g 
%7 = 10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10
? factor(g)
%8 = 
[10*x^6 + 50*x^5 - 45*x^4 - x^3 + 45*x^2 + 50*x - 10 1]

? polroots(g)
%9 = [-5.759456718855120444516616926 + 0.E-28*I, 
      0.1736274875937921035052204217 + 0.E-28*I, 
     -0.6374628782453508481344867754 - 0.5280217041759846233761928738*I,
     -0.6374628782453508481344867754 + 0.5280217041759846233761928738*I,
      0.9303774938760150186401850276 - 0.7706480276869016371542943284*I,
      0.9303774938760150186401850276 + 0.7706480276869016371542943284*I]~
? 


 x = 0.1736274875937921035052204217
? 
?  r = x * (2 + x^2)^2 * (2-5*x) /    ( (1+2*x^2)^2  * (5+2*x)  )
%12 = 0.1347361851008281269891649571
? 
? a = 1/r
%13 = 7.421911190758908573490159799
?