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.


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?


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


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.

enter image description here enter image description here enter image description here enter image description here

? 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

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:

enter image description here

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.

  • $\begingroup$ I found the coefficient with which to replace the $3$ in $-3y$ to get the two curves tangent. Came out about 7.42. There is probably some elegant way to do that, and for that matter correctly find the point of tangency that has been created this way $\endgroup$
    – Will Jagy
    Jan 28 '19 at 19:49

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