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I was reading an article on the well ordering principle and there was a problem that asked to use the well ordering principle to solve Lehman's Lemma: That there are no positive integer solutions to the following equation: $$8a^4+4b^4+2c^4=d^4$$

What I want to do is set up a counterexample set $C$ and assume that this set is non-empty. By the well ordering principle there's a least element $a\in C$. How should I use this to reach a contradiction?

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  • $\begingroup$ Perhaps you want $\,C\,$ to be the set of positive quadruples that fulfill the equation and, perhaps, also assume that $\,a\le b\le c\le d\,$...? $\endgroup$ Commented Mar 12, 2013 at 1:04
  • $\begingroup$ Yes to the first, and it says to consider the minimum value for $a$ for all possible solutions. $\endgroup$ Commented Mar 12, 2013 at 1:06

1 Answer 1

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Assume that there are positive solutions. Then there is a solution $(a,b,c,d)$ with the smallest possible value $abcd > 0$.

Considering the equation mod $2$, you see that $d$ is even, so $d = 2\delta$, and division by $2$ gives $4a^4 + 2b^4 + c^4 = 8\delta^4$. Now mod $2$ we find that $c$ is even, so $c = 2\gamma$. Going on like this, also $b = 2\beta$ and $a = 2\alpha$ are even, and we get the solution $(a/2,b/2,c/2,d/2)$ which contradicts the minimality condition on $(a,b,c,d)$.

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