Find the number of triples of real numbers $$(a,b,c)$$ such that $$a^{20}+b^{20}+c^{20}=a^{24}+b^{24}+c^{24}=1$$.

Because the exponents 20 and 24 are even, only the absolute values of the variables matter in the power sums.
Set $$|a|=x,\;|b|=y,\;|c|=z$$ with $$x,y,z\ge 0$$.

The two conditions become
$$x^{20}+y^{20}+z^{20}=1$$
$$x^{24}+y^{24}+z^{24}=1$$

Define $$f(t)=t^{24}-t^{20}=t^{20}(t^{4}-1).$$
Observe the sign of $$f(t)$$:

• If $$0\le t\le 1$$, then $$t^{4}-1\le 0$$, so $$f(t)\le 0$$.
• If $$t\gt 1$$, then $$t^{4}-1\gt 0$$, so $$f(t)\gt 0$$.

Sum the three expressions $$f(x),f(y),f(z)$$:

$$f(x)+f(y)+f(z)=\bigl(x^{24}+y^{24}+z^{24}\bigr)-\bigl(x^{20}+y^{20}+z^{20}\bigr)=1-1=0.$$

Suppose one of $$x,y,z$$ were greater than 1. Its twentieth power would already exceed 1, making $$x^{20}+y^{20}+z^{20}\gt 1,$$ contradicting the first condition. Hence every one of $$x,y,z$$ satisfies $$0\le t\le 1$$, so each $$f(t)\le 0$$.

The sum of three non-positive numbers is 0 only when each term is 0. Therefore
$$f(x)=f(y)=f(z)=0\;.$$

From $$f(t)=0$$ we get $$t^{20}(t^{4}-1)=0\implies t=0\ \text{or}\ t=1.$$
Consequently each of $$x,y,z$$ is either 0 or 1.

Now enforce $$x^{20}+y^{20}+z^{20}=1.$$ Since $$1^{20}=1$$ and $$0^{20}=0,$$ exactly one of $$x,y,z$$ equals 1 and the other two equal 0.

This means exactly one of $$|a|,|b|,|c|$$ is 1 while the others are 0. Each non-zero variable can be $$+1$$ or $$-1$$. Counting possibilities:

• Choose which position (a, b, or c) carries the ±1: 3 ways.
• Choose the sign ( + or − ) for that position: 2 ways.

Total number of ordered triples: $$3\times 2 = 6.$$

Hence, the required number of triples is
06.