# Find the solutions of the system of equations $a+b= c^n$;$b+c=a^n$;$c+a= b^n$,$\forall a,b,c \in \mathbb R$ and $n \in \mathbb Z^+$

Find the solutions of the system of equations $$a+b= c^n$$;$$b+c=a^n$$;$$c+a= b^n$$,$$\forall a,b,c \in \mathbb R$$ and $$n \in \mathbb Z^+$$

Let $$a,b$$ and $$c \in \Bbb R$$ and $$n$$ a nonnegative interger. Find all solution for the following system of equations:

$$a+b= c^n$$

$$b+c=a^n$$

$$c+a= b^n$$

Here's my solution:

Case 1: $$a \neq b \neq c$$

Subtract first equation from second, you get:

$$c-a= a^n - c^n$$ $$-1 = \frac {a^n - c^n}{a-c}$$

For $$n$$ odd, we have that if $$a>c$$, then $$a^n> c^n$$, and that division would never be negative, then $$n$$ has to be even.

$$-1 = a^{n-1} + a^{n-2}c \cdots + {a}c^{n-2}+ c^{n-1}$$

We can clearly see that at least one of them has to be negative, in fact, just one variable can be positive at most. WLOG assume $$b,c<0$$ and $$a>0$$, then:

$$b+c < 0$$ and $$a^n >0 \Rightarrow b+c \neq a^n$$ and there would be no solution. If just two are negative you can follow the same argument.

Case 2: $$a=b \neq c$$

The system will transform into:

$$2a = c^n$$ $$a+c = a^n$$

Here, we can apply the same considerations as before:

• $$n$$ is odd
• At least one variable has to be negative.

Both variables can't be negative, you can prove that by using the same argument we used in case $$1$$. Wlog, assume $$c$$ is negative.

Now:

$$a+c= a^n > 0$$ $$a> -c$$ $$\vert a \vert > \vert c\vert$$

$$\vert 2a \vert > \vert 2c\vert$$ $$\vert c^n \vert > \vert 2c\vert$$ $$\vert {c}^{n-1} \vert> 2 \Rightarrow \vert c\vert > 1$$

That's correct since $$-c>0$$

Also:

$$a+c = a^n$$ $$1+\frac {c}{a} = a^{n-1} \Rightarrow a^{n-1}<1 \Rightarrow a< 1 \Rightarrow \vert a\vert < 1$$

We have that: $$\vert a \vert > \vert c\vert$$ but $$\vert c\vert > 1$$ and $$\vert a\vert < 1$$, contradiction, hence there's no solution.

Case 3: $$a = b = c$$

The easy case:

$$2a= a^n$$ $$2=a^{n-1}$$

Then for $$n>1$$ and fixed $$n$$, we have:

$$a= {2}^{\frac {1}{n-1}}$$

For $$n=0$$, we have $$a=b=c=\frac {1}{2}$$

For $$n=1$$, we have $$a=b=c=0$$

I think the solution is complete!!, I posted it since I think it's too laborious, and I'd like to see a straightforward and elegant solution. I made some typing mistakes, can you correct them for me, please. Thanks in advance.

• I think your case 2 should be $a=b\neq c$? so that the system will transform into the two equations that you mentioned. May 24 '19 at 4:45

We divide into cases:

Suppose all three variables are negative. Then notice that $$n$$ is odd, so we can go $$a,b,c:=-a,-b,-c$$ to convert all variables to non-negatives.

Suppose exactly one variable is positive, WLOG $$a$$. Then $$a^n>0\geq b+c$$, a contradiction.

Therefore we can WLOG that $$a\geq b\geq 0$$. Since $$a^n-b^n=b-a$$ as you've proved, if $$a>b$$ then $$a^n-b^n>0>b-a$$, a contradiction. Therefore $$a=b\geq 0$$

$$2a=c^n$$. If $$c\geq0$$, we can use the same argument as above to get $$a=b=c$$, and thus $$a=0$$ or $$a=\sqrt[n-1]{2}$$

If $$c<0$$, unfortunately I can't find a simpler argument than the one you used.

Thus the only solutions are $$a=b=c$$ or $$a=b=c=\pm\sqrt[n-1]{2}$$ (where the - case only works for n odd).

• Why do you convert the variable to non-negatives. Now, for all values of $n$, $a=b=c$. That's true just if there are 3 variables. But about when there are more than 3? May 28 '19 at 14:45