Ra R.

# Show that if x,y,z are integers such that x^3 + 5y^3 = 25z^3, then x = y = z = 0.

I tried using mod 5 to get that x is divisible by 5. I don't know much about infinite descent, so I do not think that is the way to go.

The hint says "If I give you a non-trivial solution, can you find a solution which is somehow "smaller"?"

Akhil V.

Let there exist a solution to the given equation, where at least one of $x,y,z$ is not zero. We show that this is impossible, and that the only possible solution to the equation is $x=y=z=0$. CASE A: Two of $x,y,z$ are zero, but the third is not. This is clearly impossible, since the third variable has to be zero when the two zero values are plugged into the equation. CASE B: Only one of $x,y,z$ is zero. In this case, the ratio of the other two will be either $5$ or $25$. For e.g if $x=0$ but $y,z \neq 0$, then: \begin{align*} 5y^3 &= 25z^3 \\ \frac{y^3}{z^3} &= 5. \end{align*} If $y=0$, \begin{align*} x^3 &= 25z^3 \\ \frac{x^3}{z^3} &= 25. \end{align*} Since $xy,z$ are integers, the ratio of the cubes of any two of them cannot be either $5$ or $25$. Hence, this case is also impossible. CASE C: $x,y,z$ are all non-zero. Let $p,q,r$ be the exponents of $5$ in the prime factorizations of $x,y,z$ respectively. Then, the powers of $5$ in $x^3, 5y^3, 25z^3$ are $3p, 3q+1, 3r+2$. Divide the given equation by the minimum of $5^{3p}, 5^{3q+1},5^{3r+2}$. This will give a new equation, where exactly two of the three terms in the equation have $5$ as a factor. This is because since $p,q,r$ are integers, $3p \neq 3q+1 \neq 3r+2$, as they have different modulo $3$ values. Now, take modulo $5$ of the new equation. The two terms that still have at least one $5$ as a factor become zero. But then the third term must be zero modulo 5 too. However it does not have $5$ as a factor. Therefore, the term itself is zero, which is a contradiction of the assumption for this case that $x,y,z$ are all non-zero. Hence, the only possible solution to the given equation is $x=y=z=0$
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08/10/20

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