0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
0
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.