Sum is Zero, Fourth Powers Then?

given: $a + b + c = 0$ ... (G1)

given: $a^2 + b^2 + c^2 = \sqrt{74}$ ... (G2)

find: $a^4 + b^4 + c^4$

Squaring G1:

$(a+b+c)^2 = a^2 + b^2 + c^2 + 2( ab + ac + bc )$

$0 = \sqrt{74} + 2 * ( ab + ac + bc )$

$(ab + ac + bc) = -\frac{\sqrt{74})}{2}$ ... (E1)

Squaring E1:

$(ab + ac + bc)^2 = \frac{74}{4}$

$= a^2b^2 + a^2c^2 + b^2c^2 +$ $2a^2bc + 2ab^2c + 2abc^2$

$= a^2b^2 + a^2c^2 + b^2c^2 + 2abc(a + b + c)$

$= a^2b^2 + a^2c^2 + b^2c^2$

So ... $a^2b^2 + a^2c^2 + b^2c^2 = \frac{74}{4}$ ... (E2)

Squaring G2:

$(a^2 + b^2 + c^2)^2$ $= a^4 + b^4 + c^4 + 2( a^2b^2 + a^2c^2 + b^2c^2 )$

Simplifying using G2 and E2: $74 = (a^4 + b^4 + c^4 ) + \frac{74}{2}$

So: $a^4 + b^4 + c^4 = 37 \boxed{Q.E.D.}$

I suggest another approach:

Let $a=kx$ , $b=ky$ and $c=kz$ .

We see that $x+y+z=0$

Also, $k^2(x^2+y^2+z^2)=\sqrt{74}$ so $x^2+y^2+z^2=\dfrac{\sqrt{74}}{k^2}$

Seeing that we can have $x^2+y^2+z^2$ be whatever positive number we want as long as we adjust $k^2$ accordingly, we make our jobs easier by just setting simple values for $x,y,z$ . For example, let $(x,y,z)=(-1,0,1)$ .

Thus, $\dfrac{\sqrt{74}}{k^2}=2\implies k=\sqrt[4]{\dfrac{37}{2}}$

Which means $(a,b,c)=\left(-\sqrt[4]{\dfrac{37}{2}}, 0, \sqrt[4]{\dfrac{37}{2}}\right)$

Therefore $a^4+b^4+c^4=\left(-\sqrt[4]{\dfrac{37}{2}}\right)^4+(0)^4+\left(\sqrt[4]{\dfrac{37}{2}}\right)^4=\dfrac{37}{2}+0+\dfrac{37}{2}=\boxed{37}$

Another approach is to use Newton Sums: Let $P(x)=x^3+mx^2+nx+p=0$ and its roots be $a$ , $b$ and $c$ . We have then by recursion:

$S_1=a+b+c=-m$

$S_2=a^2+b^2+c^2=-mS_{1}-2n=m^2-2n$

$S_3=a^3+b^3+c^3=-mS_2-nS_1-3p=-m^3+3mn-3p$

$S_4=a^4+b^4+c^4=-mS_3-nS_2-pS_1=m^4+4mp-4m^2n+2n^2$

We know that $S_1=0$ , so $m=0$ ; and $S_2=\sqrt{74}$ , so $n=-\dfrac{\sqrt{74}}{2}$ .

Substitute them in $S_4$ and obtain, fortunately almost all terms have the factor $m$ except the last one, so:

$S_4=2n^2=2\left(-\dfrac{\sqrt{74}}{2}\right)^2=\boxed{37}$