For all prime numbers $P$ ($P > 2$), does positive rational solution $(x, y, z)$ exist for $P^x + P^y = P^z$?

Question

$\forall x,y,z \in {\mathbb Q}^+$, I want to prove that $$P^x + P^y \neq P^z$$ always holds for all prime $P$ ($P > 2$).

Answer 1

We may, and do, assume that $ x $ is an integer upon multiplication by an appropriate power of $ P $ on both sides of the equality $ P^x + P^y = P^z $.

The crucial tool we will use is the field trace. Take a Galois extension $ L/\mathbf Q $ containing all of $ P^x, P^y, P^z $. Then, we have the following result:

Lemma. Let $ P^a \in L $. Then, $ \textrm{Tr}_{L/\mathbf Q}(P^a) = 0 $ if and only if $ a $ is an integer.

Proof. Let $ a = b/c $ in reduced form, with $ c > 1 $. The trace is some positive integer multiple of the sum of all $ \mathbf Q $-conjugates of $ P^a = P^{b/c} $, and these conjugates are the $ \zeta_c^k P^{b/c} $ for $ 0 \leq k \leq c - 1 $, where $ \zeta_c $ is a primitive $ c $th root of unity. It is trivial to check that these sum to $ 0 $. On the other hand, if $ c = 1 $, then this sum is just $ P^b $, thus nonzero.

Armed with this lemma, we cover four cases:

1. Neither $ y $ nor $ z $ are integers. In this case, taking field traces on both sides of the equality yields $ P^x = 0 $, which is absurd.

2. $ y $ is an integer, but $ z $ is not. This is a contradiction, since one side of the equality $ P^x + P^y = P^z $ is then irrational, whereas the other side is an integer.

3. $ z $ is an integer, but $ y $ is not. This gives a contradiction in view of the fact that $ P^x - P^z $, being an integer, is rational; whereas $ P^y $ is not.

4. Both $ y $ and $ z $ are integers. This case has been covered by another answer; and it's a simple comparison upon assuming wlog that $ x \leq y < z $. (This is the only case for which we need the condition $ P > 2 $.)

All four cases are ruled out, thus the equation has no solutions with $ x $ integral; thus it has no (rational) solutions at all.

Answer 2

Here is a partial answer: there are no positive integral solutions if $P > 2$.

Without loss of generality, we can assume $x \le y$ so $P^x$ is a common factor of the two terms in the left hand side. Hence, $$P^x(1 + P^{y-x}) = P^z$$ If $x = y$ then the left hand side is $2P^x$ which is a multiple of $2$ and cannot be equal to $P^z$ if $P > 2$.

If $x < y$ then the left hand side is divisible by $(1 + P^{y-x})$ which is one more than a multiple of $P$ and hence not a multiple of $P$. The right hand side is only divisible by powers of $P$.

I think that I can the add the half positive integer case. I still assume $P > 2 $ and I will retain $x$, $y$, $z$ as positive integers with $x < y$ but rewrite your equation as:

$$P^{x/2} + P^{y/2} = P^{z/2}$$

Square both sides:

$$P^x + 2P^{\frac{x+y}{2}} + P^y = P^z$$

Rearranging:

$$2P^{\frac{x+y}{2}} = P^z - P^x - P^y$$

The right hand side is an integer.

If $x + y$ is even then the left hand side is also an integer but the left is even and the right is odd so that is not possible.

If $x + y$ is odd then $\frac{x + y - 1}{2}$ is an integer. So, the square root of $P$ must be rational which is not possible since P is a prime integer.

$$P^{1/2} = \frac{P^z - P^x - P^y}{2P^{\frac{x + y + 1}{2}}}$$

Further special cases may be possible in this style but will be messy. Some form of induction on the lcm of the denominators of $x, y, z$ might be possible.

Answer 3

@badjohn, the fact that $P^x + P^y = P^z$ has no solutions in integers is trivial isn't it? (how can odd number + odd number = odd number?)

Answer 4

Here's an answer with no Galois theory.

Suppose wlog that $ x \leq y \lt z $ are (not necessarily positive) rationals solving the equation $P^x + P^y = P^z$ for some prime $P$.

Then divide by $P^x$ to obtain $1 + P^{y-x} = P^{z-x}$.

Now write $y - x = \frac{a}{d}$ and $z - x = \frac{b}{d}$ for some integers $a \geq 0, b \gt 0, d \gt 0$. Moreover, $b > a$.

Now $P^{\frac{1}{d}}$ is a common root of $X^d - P$ and $X^b - X^a - 1$. Since $X^d - P$ is irreducible (by Eisenstein's criterion), this implies that $X^d - P$ divides $X^b - X^a - 1$.

Thus $X^b - X^a - 1 = f(X)(X^d - P)$ for some polynomial $f(X)$ which can be assumed to be in $\mathbb{Z}[X]$ by Gauss' lemma. But, if $a > 0$ then the constant term on the RHS is a multiple of $P$, which contradicts the constant term of $-1$ on the LHS. Thus $a = 0$, and $P = 2$.

Translating back to our old variables, this means $x = y$ and we get $2^x + 2^x = 2^z$ with only solution $y = x, z = x + 1, P = 2$.