Show that solution $(0,0)$ is unique Assume the system:
$$
\begin{align*}
&amp; 2x-5y-x(x^2+2y^2)^2=0 \\
&amp; 5x+2y-3y(2x^2+y^2)^2=0 
\end{align*}
$$
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?
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,
$$5t^6-22t^5+80t^4-16t^3+80t^2+2t+15=$$
$$=(5t^6-22t^5+25t^4)+(55t^4-16t^3+40t^2)+(40t^2+2t+15)&gt;0.$$
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
$$ \eqalign{ -x &amp;\left( 6561\,{s}^{4}{x}^{24}-18954\,{s}^{4}{x}^{20}-23328\,{s}^{3}{
x}^{20}+21708\,{s}^{4}{x}^{16}+173340\,{s}^{3}{x}^{16}\right.\cr &amp;+160704\,{s}^{2}
{x}^{16}-23976\,{s}^{4}{x}^{12}-191016\,{s}^{3}{x}^{12}-396000\,{s}^{2
}{x}^{12}\cr &amp;
-248832\,s{x}^{12}+10692\,{s}^{4}{x}^{8}+24540\,{s}^{3}{x}^{8
}+597520\,{s}^{2}{x}^{8}+1517568\,s{x}^{8}-5832\,{s}^{4}{x}^{4}
\cr &amp;\left.+746496
\,{x}^{8}-14938\,{s}^{3}{x}^{4}+47328\,{s}^{2}{x}^{4}+337792\,s{x}^{4}
-59392\,{x}^{4}+18125\,{s}^{3} \right)}
$$
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.