Gorgeous diophantine equation

Question

How to find all integer solutions of the following equation? $$y^7=14 \cdot 3^{100}x^6 + 70 \cdot 3^{300} x^4 + 42 \cdot 3^{500} x^2 + 2 \cdot 3^{700}$$

Answer 1

Assume $x=3^{100} a$ to balance all those large powers of 3:

$$ y^7 = 2\cdot 3^{700} \left( 7a^6 + 35a^4 + 21a^2 + 1 \right) $$

Due to the common factor, $y$ should be of the form $y=2\cdot 3^{100} b$:

\begin{align} 64 b^7 &= 7a^6 + 35a^4 + 21a^2 + 1 \\ &= \sum_{k=0}^3 \binom7{2k} a^{2k} \\ &= \frac{(1+a)^7 + (1-a)^7}2 \\ (2b)^7 &= (1+a)^7 + (1-a)^7 \end{align}

The last equation only allows solutions of the form $a=\pm 1,b=1$, so two sets of solutions are:

$$ x = \pm3^{100},\quad y = 2\cdot 3^{100} $$

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We don't really need to assume $3^{100}\mid x$. Following the idea above original equation can be rewritten as

\begin{align} y^7 &= 2 \cdot \sum_{k=0}^3 \binom{7}{2k} 3^{100(7-2k)}x^{2k} \\ &= (3^{100} - x)^7 + (3^{100} + x)^7 \end{align}

which does not have any integer solutions except $x=\pm 3^{100}$ (Fermat's last theorem, $n=7$). So the above two are the only solutions.