# Number of solutions in degree four

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.

• Assuming that you gave the KVPY exam, the question asked for number of “real” solutions, not integral ones – N.S.JOHN Nov 4 '18 at 18:25
• i could solve for it. i just wanted to change the question to integer solutions. – maveric Nov 4 '18 at 20:52

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.

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.

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.