Systems of Nonlinear Equations: Very Difficult Problems with Solutions

Solution:
Treat [tex]z[/tex] as known. From the system, [tex]x+y=4-z[/tex] and [tex]xy=\frac{16+z^2}{2}[/tex], hence
[tex](x-y)^2=(x+y)^2-4xy=(4-z)^2-2\left(16+z^2\right)=-z^2-8z-16=-(z+4)^2[/tex]
The left-hand side is [tex]\ge 0[/tex] and the right-hand side is [tex]\le 0[/tex], so both are equal to [tex]0[/tex]:
[tex]z=-4,\quad x=y,\quad x+y=4-z=8\Rightarrow x=y=4[/tex]
Check: [tex]4+4-4=4[/tex] and [tex]2\times 4\times 4-(-4)^2=32-16=16[/tex]. Although there are only two equations for three unknowns, the system has exactly one real solution: [tex]\left(4, 4, -4\right)[/tex].

Solution:
Write the equations as
[tex]a^{3}-3ab^{2}=11,\qquad 3a^{2}b-b^{3}=-2[/tex]
Square both equations and add them. On the left we get
[tex]\left(a^{3}-3ab^{2}\right)^{2}+\left(3a^{2}b-b^{3}\right)^{2}=a^{6}+3a^{4}b^{2}+3a^{2}b^{4}+b^{6}=\left(a^{2}+b^{2}\right)^{3}[/tex]
and on the right [tex]11^{2}+(-2)^{2}=125[/tex]. Since [tex]a^{2}+b^{2}[/tex] is a real number,
[tex]\left(a^{2}+b^{2}\right)^{3}=125\Rightarrow a^{2}+b^{2}=5[/tex]
Such numbers exist, for example [tex]a=-1[/tex], [tex]b=2[/tex]: [tex]-1=3\times (-1)\times 4+11[/tex] and [tex]8=3\times 1\times 2+2[/tex].
Remark: this is the identity [tex]\left|(a+bi)^{3}\right|^{2}=\left(a^{2}+b^{2}\right)^{3}[/tex], because [tex](a+bi)^{3}=\left(a^{3}-3ab^{2}\right)+\left(3a^{2}b-b^{3}\right)i[/tex].

Solution:
Add the first and the third equations and subtract twice the second:
[tex]\left(x^6-2x^5+x^4\right)+\left(y^6-2y^5+y^4\right)+\left(z^6-2z^5+z^4\right)=3-6+3=0[/tex]
For every real [tex]t[/tex]
[tex]t^6-2t^5+t^4=t^4(t-1)^2\ge 0[/tex]
A sum of three non-negative numbers equals [tex]0[/tex] only if each of them is [tex]0[/tex], so [tex]x^4(x-1)^2=y^4(y-1)^2=z^4(z-1)^2=0[/tex], i.e. each of [tex]x,y,z[/tex] equals [tex]0[/tex] or [tex]1[/tex].
Then [tex]x^4+y^4+z^4[/tex] equals the number of unknowns that are equal to [tex]1[/tex]. Since this sum is [tex]3[/tex], all three are equal to [tex]1[/tex].
Conversely, [tex]x=y=z=1[/tex] satisfies all three equations, so [tex]\left(1, 1, 1\right)[/tex] is the only real solution.

Solution:
Adding and subtracting the equations gives an equivalent system:
[tex]\left\{\begin{array}{l} x^3+y^3=7(x+y) \\ x^3-y^3=3(x-y) \end{array}\right.\iff\left\{\begin{array}{l} (x+y)\left(x^2-xy+y^2-7\right)=0 \\ (x-y)\left(x^2+xy+y^2-3\right)=0 \end{array}\right.[/tex]
We combine one factor from each equation. Case [tex]1[/tex]: [tex]x+y=0[/tex] and [tex]x-y=0[/tex], which gives [tex]\left(0, 0\right)[/tex].
Case [tex]2[/tex]: [tex]x+y=0[/tex] and [tex]x^2+xy+y^2=3[/tex]. With [tex]y=-x[/tex] we get [tex]x^2=3[/tex]: [tex]\left(\sqrt{3}, -\sqrt{3}\right)[/tex] and [tex]\left(-\sqrt{3}, \sqrt{3}\right)[/tex].
Case [tex]3[/tex]: [tex]x-y=0[/tex] and [tex]x^2-xy+y^2=7[/tex]. Then [tex]x^2=7[/tex]: [tex]\left(\sqrt{7}, \sqrt{7}\right)[/tex] and [tex]\left(-\sqrt{7}, -\sqrt{7}\right)[/tex].
Case [tex]4[/tex]: [tex]x^2-xy+y^2=7[/tex] and [tex]x^2+xy+y^2=3[/tex]. Then [tex]xy=-2[/tex], [tex]x^2+y^2=5[/tex] and [tex](x+y)^2=1[/tex].
If [tex]x+y=1[/tex], then [tex]x,y[/tex] are the roots of [tex]t^2-t-2=0[/tex]: [tex]\left(2, -1\right)[/tex] and [tex]\left(-1, 2\right)[/tex]. If [tex]x+y=-1[/tex], they are the roots of [tex]t^2+t-2=0[/tex]: [tex]\left(1, -2\right)[/tex] and [tex]\left(-2, 1\right)[/tex].
The cases do not overlap, because [tex]xy[/tex] equals [tex]0[/tex], [tex]-3[/tex], [tex]7[/tex] and [tex]-2[/tex] in cases [tex]1[/tex], [tex]2[/tex], [tex]3[/tex], [tex]4[/tex] respectively. Hence the number of solutions is
[tex]1+2+2+4=9[/tex]

Solution:
The left-hand sides are non-negative, hence so are the right-hand sides: [tex]x\ge 0[/tex], [tex]y\ge 0[/tex], [tex]z\ge 0[/tex]. For every [tex]t\ge 0[/tex]
[tex]t-\frac{2t^2}{1+t^2}=\frac{t+t^3-2t^2}{1+t^2}=\frac{t(t-1)^2}{1+t^2}\ge 0[/tex]
i.e. [tex]\frac{2t^2}{1+t^2}\le t[/tex], with equality only for [tex]t=0[/tex] or [tex]t=1[/tex]. Applying this with [tex]t=x[/tex], [tex]t=y[/tex], [tex]t=z[/tex] to the three equations gives [tex]y\le x[/tex], [tex]z\le y[/tex], [tex]x\le z[/tex]. Hence
[tex]x\le z\le y\le x\Rightarrow x=y=z[/tex]
Then the first equation becomes [tex]\frac{2x^2}{1+x^2}=x[/tex], i.e. [tex]x(x-1)^2=0[/tex], so [tex]x=0[/tex] or [tex]x=1[/tex]. Both triples are solutions: [tex]\frac{2\times 0}{1+0}=0[/tex] and [tex]\frac{2\times 1}{1+1}=1[/tex]. Therefore the only real solutions are [tex]\left(0, 0, 0\right)[/tex] and [tex]\left(1, 1, 1\right)[/tex].

Solution:
Square the first equation and subtract the second:
[tex]6xy=(\alpha+2)^2-(5\alpha-2)=\alpha^2-\alpha+6\Rightarrow xy=\frac{\alpha^2-\alpha+6}{6}[/tex]
Now we find for which [tex]\alpha[/tex] the system has real solutions. Put [tex]u=3x[/tex], [tex]v=y[/tex]; then [tex]u+v=\alpha+2[/tex] and [tex]uv=3xy=\frac{\alpha^2-\alpha+6}{2}[/tex]. Real [tex]u,v[/tex] exist exactly when the quadratic equation [tex]t^2-(\alpha+2)t+\frac{\alpha^2-\alpha+6}{2}=0[/tex] has real roots:
[tex](\alpha+2)^2-2\left(\alpha^2-\alpha+6\right)\ge 0\iff \alpha^2-6\alpha+8\le 0\iff \alpha\in\left[2, 4\right][/tex]
Conversely, for such [tex]\alpha[/tex] these [tex]u,v[/tex] satisfy [tex]u+v=\alpha+2[/tex] and [tex]u^2+v^2=(u+v)^2-2uv=5\alpha-2[/tex], so they give a solution of the system.
The quadratic function [tex]f(\alpha)=\frac{\alpha^2-\alpha+6}{6}[/tex] has a positive leading coefficient and its vertex at [tex]\alpha=\frac{1}{2}[/tex], so it is increasing on [tex]\left[2, 4\right][/tex]. Its greatest value is attained at [tex]\alpha=4[/tex], where [tex]xy=3[/tex] (the solution is [tex]x=1[/tex], [tex]y=3[/tex]).

Solution:
Domain: [tex]x\gt 0[/tex], [tex]x+y\gt 0[/tex]. The first equation is [tex]\log_2(x+y)=\log_2(2x)[/tex], i.e. [tex]x+y=2x[/tex], so [tex]y=x[/tex] (then [tex]x+y=2x\gt 0[/tex] holds automatically), and the second equation becomes [tex](x-a)^2+a^2=1[/tex]. The system is equivalent to
[tex]\left\{\begin{array}{l} y=x \\ x\gt 0 \\ (x-a)^2=1-a^2 \end{array}\right.[/tex]
so the equation [tex](x-a)^2=1-a^2[/tex] must have exactly one positive root. If [tex]|a|\gt 1[/tex], there are no roots. If [tex]a=\pm 1[/tex], the only root is [tex]x=a[/tex]; it is positive only for [tex]a=1[/tex], so [tex]a=1[/tex] works and [tex]a=-1[/tex] does not.
If [tex]|a|\lt 1[/tex], there are two different roots [tex]x_{1,2}=a\pm\sqrt{1-a^2}[/tex], and both conditions must hold:
[tex]a+\sqrt{1-a^2}\gt 0,\qquad a-\sqrt{1-a^2}\le 0[/tex]
The first holds for [tex]a\ge 0[/tex]; for [tex]a\lt 0[/tex] it means [tex]1-a^2\gt a^2[/tex], i.e. [tex]a\gt -\frac{\sqrt{2}}{2}[/tex]. The second holds for [tex]a\le 0[/tex]; for [tex]a\gt 0[/tex] it means [tex]a^2\le 1-a^2[/tex], i.e. [tex]a\le\frac{\sqrt{2}}{2}[/tex]. So [tex]-\frac{\sqrt{2}}{2}\lt a\le\frac{\sqrt{2}}{2}[/tex].
At the endpoints: for [tex]a=\frac{\sqrt{2}}{2}[/tex] the roots are [tex]\sqrt{2}[/tex] and [tex]0[/tex], and only [tex]x=\sqrt{2}[/tex] is in the domain; for [tex]a=-\frac{\sqrt{2}}{2}[/tex] the roots are [tex]0[/tex] and [tex]-\sqrt{2}[/tex], so there is no solution. Therefore
[tex]a\in\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right]\cup\left\{1\right\}[/tex]

Solution:
Write the equations as [tex]y\left(1-x^{2}\right)=2x[/tex], [tex]z\left(1-y^{2}\right)=2y[/tex], [tex]x\left(1-z^{2}\right)=2z[/tex]. If [tex]x^{2}=1[/tex], the first equation gives [tex]2x=0[/tex], a contradiction; in the same way [tex]y^{2}\ne 1[/tex] and [tex]z^{2}\ne 1[/tex]. Hence the system is equivalent to
[tex]y=\frac{2x}{1-x^{2}},\qquad z=\frac{2y}{1-y^{2}},\qquad x=\frac{2z}{1-z^{2}}[/tex]
Every real number [tex]x[/tex] can be written uniquely as [tex]x=\tan u[/tex] with [tex]u\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex]. Since [tex]x\ne\pm 1[/tex], the double-angle formula gives [tex]y=\frac{2\tan u}{1-\tan ^{2}u}=\tan 2u[/tex]; in the same way, since [tex]y\ne\pm 1[/tex] and [tex]z\ne\pm 1[/tex],
[tex]z=\tan 4u,\qquad x=\tan 8u[/tex]
So [tex]\tan 8u=\tan u[/tex], i.e. [tex]8u=u+k\pi[/tex], [tex]u=\frac{k\pi}{7}[/tex], [tex]k\in\mathbb{Z}[/tex]. In the interval [tex]\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex] this gives [tex]k=-3,-2,\ldots,3[/tex]: [tex]7[/tex] values of [tex]u[/tex], hence [tex]7[/tex] different values of [tex]x[/tex], and each of them determines [tex]y[/tex] and [tex]z[/tex] uniquely.
Conversely, let [tex]u=\frac{k\pi}{7}[/tex]. A number [tex]\frac{j\pi}{7}[/tex] with [tex]j\in\mathbb{Z}[/tex] is never equal to [tex]\frac{\pi}{2}+m\pi[/tex] or [tex]\frac{\pi}{4}+\frac{m\pi}{2}[/tex] with [tex]m\in\mathbb{Z}[/tex], because that would mean [tex]2j=7(2m+1)[/tex] or [tex]4j=7(2m+1)[/tex], an even number equal to an odd one. Applying this to [tex]j=k,\ 2k,\ 4k[/tex], we see that [tex]x=\tan u[/tex], [tex]y=\tan 2u[/tex], [tex]z=\tan 4u[/tex] are defined and none of them equals [tex]\pm 1[/tex].
The first two equations hold by the double-angle formula, and the third holds because
[tex]\frac{2\tan 4u}{1-\tan ^{2}4u}=\tan 8u=\tan \left(u+k\pi\right)=\tan u[/tex]
Thus the system has exactly [tex]7[/tex] real solutions, for example [tex]\left(0, 0, 0\right)[/tex] and [tex]\left(\tan \frac{\pi}{7}, \tan \frac{2\pi}{7}, \tan \frac{4\pi}{7}\right)[/tex].