Talk:Jacobi–Madden equation

Restrictions and Possibilities on the Numbers

Obviously, the equation is true if a = b = c = d = 0. Also, in the set { a, b, c, d }, if three of the four are zero, then the equation is true. If two of the four are zero, then we get $a^{4}+b^{4}=(a+b)^{4}$, and Pierre de Fermat, himself, showed that this one is impossible for all nonzero numbers { a, b, c }, with $a^{4}+b^{4}=c^{4}$.

If one of the numbers is zero, then we get $a^{4}+b^{4}+c^{4}=(a+b+c)^{4}$. This is obviously not true for positive integers, because the right-hand side of the equation would be too large. Possibly,it might be true if one of them is allowed to be negative.

If none of the numbers is zero, then we get $a^{4}+b^{4}+c^{4}+d^{4}=(a+b+c+d)^{4}$. This is also obviously not true for positive integers, because the right-hand side of the equation would be too large. It has now been shown that there are infinitely-many solutions if some of the numbers are allowed to be negative, and some positive.

This is the new section that I have put in today.74.163.40.58 (talk) 19:41, 29 September 2008 (UTC)

Hi 74.163.40.58, I removed this section. If you disagree, just put it back and I won't remove it again. I'll explain you why: as you say yourself, the restrictions on the solutions you mention are rather obvious, and don't give any extra insight into the problem. Even the observation that there are no solutions with c and d equal to 0 is very easy, you don't need Fermat's result and can just remark that a solution (a,b) with a and b coprime is equal to a zero of $2a^{2}+3ab+2b^{2}$ (which is the difference between the LHS and the RHS divided by $2ab$), which doesn't have nonzero solutions modulo 3. Doetoe (talk) 11:31, 24 October 2008 (UTC)

I removed the text about Euler's sum of powers conjecture, since that already has its own page. Doetoe (talk) 11:31, 24 October 2008 (UTC)