Do you know Eric Lehman?

Suppose that there exists a solution. Then there must be a solution in which a has the smallest possible value. We will show that, in this solution, a, b, c, and d must all be even. How­ ever, we can then obtain another solution over the positive integers with a smaller a by dividing a, b, c, and d in half. This is a contradiction, and so no solution exists. All that remains is to show that a, b, c, and d must all be even. The left side of Lehman’s equation is even, so $d^4$ is even, so d must be even. Substituting $d = 2d^{'}$ into Lehman’s equation gives:

$\large{8a^4 + 4b^4 + 2c^4 = (d^{'})^4 }$

Now $2c^4$ must be a multiple of 4, since every other term is a multiple of 4. This implies that $c^4$ is even and so c is also even. Substituting $c = 2c^{'}$ gives

$\large{8a^4 + 4b^4 + 32(c^{'})^4 = (d^{'})^4 }$

Arguing in the same way, $4b^4$ must be a mutliple of 8, since every other term is. Therefore, $b^4$ is even and so b is even. Substituting $b=2b^{'}$ gives:

$\large{8a^4 + 64(b^{'})^4 + 32(c^{'})^4 = (d^{'})^4 }$

Finally, $8a^4$ must be a multiple of 16, $a^4$ must be even, and so a must also be even. Therefore, a, b, c, and d must all be even, as claimed.