Triplets 4th of power of first equals to sum of other two

Question

I was looking through the admission test for the University Normale of Pisa and I found a problem that I don't know how to solve, it state something like that:

Find all the triplets of number (x, y, z) such that: $$ \left\{ \begin{array}\ x^4 = y + z\\ y^4 = z + x\\ z^4 = y + x \end{array} \right. $$

EDIT: It ask for real number.

All: Please observe that the question was earlier tagged elementary-number-theory. This lead several posters to assume that the variables should be integers. Several answers may appear to be strange as a consequence of that./JL

Answer 1

1 ~ [y + z] mod(5) and 1 ~ [z + x] mod(5) and 1 ~ [y + y] mod(5) from fermat's little theorem. Then since 5 is prime we know that each inverse must be unique. Thus this suggests that y mod(5) = z mod(5) = x mod(5) = 3 mod(5) As for finding all specific solutions I'm not sure.

Edit: I now see that you changed it to [y + x] so x,y,z it shouldn't be equivalent to 3 mod(5) as I said above, but then x, y are inverses mod (5) and so are y,z and z,x

Answer 2

Without loss of generality we can assume that $x \geq y \geq z$. Then it is clear that $x \geq 0$ (otherwise all three are negative, which is not possible)

$$x^4=y+z \le 2x \,.$$

Thus $x^3 \leq 2 \Rightarrow x=0 \, \mbox{or} \, x=1 \,.$

If $x=0$ we have $y,z \leq 0$ and $y=z^4 \geq 0 \Rightarrow y=0$ and $z=0$.

If $x=1$, then $z \leq y \leq 1$ and $y+z=1$. This implies that $z=0$ and $y=1$. But this doesn't work.

Thus, there is only 1 solution $(0,0,0)$.

Answer 3

$$y^4-x^4=x-y$$

If $x\ne y$

$$(y^2+x^2)(y+x)=-1$$

Assuming that last line should be $z^4=x+y$ we have

$$(x^2+y^2)z^4=-1$$

which has no solution. Therefore $x$ must equal $y$. Similarly, we can get $y=z$. So we have

$$x^4=2x$$

The only real solutions are $x,y,$ and $z$ all equal $0$ or the cube root of $2$.