<s>[INST] Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$. Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.
I have shown it is not possible for $a=4$, but not for any a greater than $4$.
 [/INST] We see that
$$\frac{x+y}{xy}=\frac{2^a}{7^a}$$
and that leads to
$$7^a(x+y)=2^a(xy)$$
now let $\gcd(x,y)=d$, and write $x=dx_0$ and $y=dy_0$.
We get:
$$7^a(dx_0+dy_0)=2^a(dx_0dy_0)$$
or, rewritten:
$$7^a(x_0+y_0)=2^a(dx_0y_0)$$
note that since $\gcd(x_0,y_0)=1$, we know $\gcd(x_0y_0,x_0+y_0)=1$; thus, $x_0$ and $y_0$ can only contain factors $7$ (since $x_0y_0|7^a(x_0+y_0)$ implies $x_0y_0|7^a$). Now write $x=dx_0=d\cdot7^n$ and $y=dy_0=d\cdot7^m$. We substitute this into the equation to get
$$7^a(7^n+7^m)=2^a(d\cdot7^{n+m})$$
meaning, $2^a|7^n+7^m$. Now set w.l.o.g. $n\geq m$, so that we can write $7^n+7^m=7^m(7^{n-m}+1)$. We know there's no factors $2$ in $7^m$, so we can say $2^a|7^{n-m}+1$. We can write $n-m=k$ to simplify this to $2^a|7^k+1$. Proving that $7^k+1$ can never contain four factors $2$ would thus be sufficient.
Let's look at this $\mod 16$. We see that $7^2=49\equiv 1\mod 16$, and as such, we know that $7^k+1\equiv 2\mod 16$ if $k$ is even, and if $k$ is odd, $7^k+1=7\cdot 7^{k-1}+1\equiv 7\cdot 1+1\equiv 8\mod 16$. Therefore, $7^k+1$ is never divisible by $16$, and as a result of that, never contains 4 or more prime factors. This proves that the initial equality has no integer solutions if $a>3$.
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