Tomi's Challenges - #8

$\Large e^{t^2-ty^7}e^{te^{t-y^7}}=\left(\frac{x^7e^{x^7}+y^7e^{x^7}}{t}\right)^t\cdot e^{x^7e^{x^7}+y^7e^{x^7}}$

After taking the $t$ -th root on both sides, we get:

$\Large e^{t-y^7}e^{e^{t-y^7}}=\left(\frac{x^7e^{x^7}+y^7e^{x^7}}{t}\right)\cdot e^{\frac{x^7e^{x^7}+y^7e^{x^7}}{t}}$

Taking the Lambert $W$ function on both sides, we simplify this equation drastically.

$\Large W\left(e^{t-y^7}e^{e^{t-y^7}}\right)=W\left(\left(\frac{x^7e^{x^7}+y^7e^{x^7}}{t}\right)\cdot e^{\frac{x^7e^{x^7}+y^7e^{x^7}}{t}}\right)$

$\Large e^{t-y^7}=\frac{x^7e^{x^7}+y^7e^{x^7}}{t}$

$\Large \frac{e^{t}}{e^{y^7}}=\frac{e^{x^7}(x^7+y^7)}{t}$

$\Large \frac{te^{t}}{e^{y^7}}=e^{x^7}(x^7+y^7)$

$te^{t}=e^{x^7}e^{y^7}(x^7+y^7)$

$W(te^{t})=W(e^{x^7+y^7}(x^7+y^7))$

$t=x^7+y^7$

Now we just need to show that $\sqrt[7]{t}$ is an integer. We can easily see that $t$ is divisible by $10^7$ . It can be also shown that $2187=3^7$ , by trying to divide by $3$ multiple times. So, we have:

$30^7=x^7+y^7$ , which has no solutions by Fermat's last theorem .