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:

$$a+b= c^n$$


$$ c+a= b^n $$

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.


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

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

That's correct since $-c>0$


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

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.

  • $\begingroup$ I think your case 2 should be $a=b\neq c$? so that the system will transform into the two equations that you mentioned. $\endgroup$ 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).

  • $\begingroup$ 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? $\endgroup$
    – Vmimi
    May 28 '19 at 14:45

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