Fall Mathematical Competition

Let $a+b+c=0$. Then $c=-a-b$.

Substitute into $a^4+b^4+c^4$: $$a^4+b^4+(-a-b{)}^4=50$$ Expand $(-a-b{)}^4$: $$(-a-b{)}^4=(a+b{)}^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$$ So: $$a^4+b^4+[a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4]=50$$ $$2a^4+2b^4+4a^{3}b+4a{b}^3+6a^2{b}^2=50$$ Group terms: $$2(a^4+b^4)+4(a^{3}b+a{b}^3)+6a^2{b}^2=50$$ Note that $a^{3}b+a{b}^3=ab(a^2+b^2)$.

Let $s=a+b$, $p=ab$. Since $a+b+c=0$, $c=-s$.

Now, $a^2+b^2=(a+b{)}^2-2ab=s^2-2p$.

Also, $a^4+b^4=(a^2{)}^2+(b^2{)}^2=[a^2+b^2{]}^2-2a^2{b}^2=(s^2-2p{)}^2-2p^2$.

$a^2{b}^2=(ab{)}^2=p^2$.

Substitute into the equation: $$2[(s^2-2p{)}^2-2p^2]+4p(s^2-2p)+6p^2=50$$ Expand $(s^2-2p{)}^2$: $$(s^2-2p{)}^2=s^4-4s^{2}p+4p^2$$ So: $$2[s^4-4s^{2}p+4p^2-2p^2]+4p(s^2-2p)+6p^2=50$$ $$2[s^4-4s^{2}p+2p^2]+4p{s}^2-8p^2+6p^2=50$$ $$2s^4-8s^{2}p+4p^2+4p{s}^2-8p^2+6p^2=50$$ $$2s^4-8s^{2}p+4p{s}^2+(4p^2-8p^2+6p^2)=50$$ $$2s^4-8s^{2}p+4p{s}^2+2p^2=50$$ Recall $s=a+b$, but $a+b+c=0$, so $s=-c$. But $a$, $b$, $c$ are symmetric, so $s$ can be any real number such that $a+b+c=0$.

But $a+b+c=0$ implies $s+c=0$, so $c=-s$.

Now, $ab+bc+ca=ab+b(-s)+a(-s)=ab-bs-as=ab-s(a+b)=ab-s^2$. But $a+b=s$, so $ab+bc+ca=ab-s^2$.

Let $x=ab+bc+ca=ab-s^2$.

But $ab=x+s^2$.

Let us try to find $x$.

Let us try specific values. Since $a+b+c=0$, let $a=x$, $b=y$, $c=-x-y$.

Then $a^4+b^4+c^4=x^4+y^4+(-x-y{)}^4=50$.

Expand $(-x-y{)}^4$: $$(-x-y{)}^4=(x+y{)}^4=x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4$$ So: $$x^4+y^4+[x^4+4x^{3}y+6x^2{y}^2+4x{y}^3+y^4]=50$$ $$2x^4+2y^4+4x^{3}y+4x{y}^3+6x^2{y}^2=50$$ Let $x=t$, $y=-t$, $c=0$. Then $a+b+c=t+(-t)+0=0$.

$a^4+b^4+c^4=t^4+(-t{)}^4+0^4=2t^4$. Set $2t^4=50$, $t^4=25$, $t=\pm \sqrt{5}$.

So $a=\sqrt{5}$, $b=-\sqrt{5}$, $c=0$.

Then $ab+bc+ca=(\sqrt{5})(-\sqrt{5})+(-\sqrt{5})(0)+(0)(\sqrt{5})=-5+0+0=-5$.

Thus, $ab+bc+ca=-5$.