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Find number of postive integra solutions of the equation $ x^4+ 4y^4 + 16z^4 +64= 32xyz$.

I could just proceed till that x cant be odd.

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  • $\begingroup$ Assuming that you gave the KVPY exam, the question asked for number of “real” solutions, not integral ones $\endgroup$ – N.S.JOHN Nov 4 '18 at 18:25
  • $\begingroup$ i could solve for it. i just wanted to change the question to integer solutions. $\endgroup$ – maveric Nov 4 '18 at 20:52
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There are no integer solutions.

As you remark, $x$ must be even. Write $x=2x_1$. Then we see $$16x_1^4+4y^4+16z^4+64=64x_1yz\implies 16\,|\,4y^4\implies y=2y_1$$

Continuing we get $$4x_1^4+16y_1^4+4z^4+16=16x_1y_1z\implies x_1^4+4y_1^4+z^4+4=4x_1y_1z$$

Thus, $x_1\equiv z\pmod 2$. If they were both odd, we'd have $x_1^4,z^4\equiv 1 \pmod 4$ in which case out last equation becomes $2\equiv 0 \pmod 4$, a contradiction. Thus they are both even. Write $x_1=2x_2, z=2z_1$. We have $$16x_2^4+4y_1^4+16z_1^4+4=16x_2y_1z_1\implies 4x_2^4+y_1^4+4z_1^4+1=4x_2y_1z_1$$

Thus $y_1$ must be odd. But in that case $y_1^4\equiv 1\pmod 4$ in which case our last equation becomes $2\equiv 0 \pmod 4$, a contradiction, and we are done.

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There do not exist any positive integral solutions to this equation. Simply apply the AM-GM inequality to the LHS to arrive at this conclusion.

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AM-GM inquelity $$ \\x^4+4y^4+16z^4+64\geq 4{(x^4(4y^4)(16z^4)(64))}^{1/4}=32|xyz|\geq 32xyz=x^4+4y^4+16z^4+32 $$ this inquelity is equal if and only if $x^4=4y^4=16z^4=64$ and $xyz\geq0$ but $x^4=4y^4=>|x|=4^{1/4}|y|=>x=y=0=>z=0$ $=>$ this question no positive integer soluation. $$\\$$Sorry for my english.

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