Here is my solution's Sketch

Uh uhhh and a recycled problem again

Thailand MO 2018 P6

Iora wrote: If $x,y,z \in\mathbb{N}$ and $2x^2+3y^3=4z^4$, Prove that $6|x,y,z$ So easy problem just look first mod 3 and u will see that 3 divides x so 3 also divides z.Then again looking at mod 3 and mod 2 we get the result

you can also just take everything mod 6

Easy peasy for olympiads its enough for look at module{2,3,4,6}

We will prove that $x, y, z$ are divisible by both 2 and 3. By mod 2: $y \equiv 0 $ (mod 2) Looking mod 3: $2x^2 + 3y^3 \equiv 2$ (mod 3) if x is not divisble by 3 and 0 otherwise. On the other hand, $4z^4 \equiv 1$ (mod 3) if z is divisble by 3 and 0 in the other case. We conclude that $x, z$ can be divided by 3. By mod 2: $y$ is even. By mod 4: $x$ is even, so $x$ is divisible by 6 $...(1)$ By mod 8: $z$ is even, so $z$ is divisible by 6. $...(2)$ By mod 9: $2x^2 \equiv 2 * 6^2 \equiv 0$ (mod 3) and $4y^4 \equiv 4 * (6^2)^2 \equiv 0$ (mod 9), so $3y^3$ is divisble by $9$. This means that $y^3$ is divisble by $9$, hence $y$ is divisible by $3$. Furthermore, $y$ is divisble by $6$. $...(3)$ From $(1)$, $(2)$ and $(3)$, we can conclude that $6|x, y, z$. Huge thanks to @lora for the hint!

@above just check $\mod 9$

It is a very intuitive question, even for a junior it is very easy

Pretty easy, just take mod 2, mod 3, 4, 8, 9 (that's what I did)

Looking this equation for mod 3 and 2 .

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$For mod 2; 3y^3==y^3==0 (mod 3)$

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$ So; y=3k; Looking for mod 3; 2x^2==z^4 (mod 3) (z^2)^2== 0,1 (mod 3) So, x=3a;$ Let's write x=3a and y=3k instead of x,y in our equation

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$ 18a^2+81k^3=4z^4$

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$3*(6a^2+27k^3)=4z^4$

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$4z^4$ that's divisible by 2, so, the left side must also divided into 2. So if both of them are divided into 2 and 3, that means, they are also divided into 6. If the left handside divisible by 6, so right handside must also; So;

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$6/x,y,z$