Consider the Diophantine equation:$$a^4+b^4+c^4+d^4=(a+b+c+d)^4$$It's clear that no set of positive numbers will satisfy this equation, at least one of the numbers will need to be negative. It's also clear that if we find a solution \(n=a+b+c+d\) then this will generate an infinity of solutions between all multiples of this number will satisfy the equation:$$ \begin{align} (na)^4+(nb)^4+(nc)^4+(nd)^4 &= (n *(a+b+c+d))^4\\n^4 a^4 +n^4 b^4 + n^4 c^4 + n^4 d^4 &= n^4 (a+b+c+d)^4 \\a^4+b^4+c^4+d^4 &= (a+b+c+d)^4 \end{align} $$where \(n\) can be any integer, positive or negative, or even zero. So what is the first number that satisfies the equation? In 1964, somebody named Brudno found the number 5491 that satisfies where 5491 is written as 955 + 1700 - 2364 + 5400 and we have:$$955^4+1700^4+(-2364)^4+5400^4\\=(955 + 1700 - 2364 + 5400)^4$$In terms of positive numbers, this means that the progressive multiples of 5491 are also solutions viz. 10982, 16473, 21964, 27455, 32946, 38437, 43928, 49419, ... etc..
Are there other numbers? Well somebody named Wroblewski found 51361 to be a solution when it written as 48150 - 31764 + 27385 + 7590. Thus:$$48150^4 +(-31764)^4+27385^$+7590^4\\=(48150 - 31764 + 27385 + 7590)^4$$At this point, it's relevant to include this quote from the comments to OEIS A138760: numbers \(n\) such that \(n^4\) is a sum of 4th powers of four nonzero integers whose sum is \(n\):
Any multiple of a member is also a member. A member that is not a multiple of another member is called primitive. Using elliptic curves, Jacobi and Madden prove that there are infinitely many primitive members. According to them, the only primitive members less than 222,000 are 5491 (due to Brudno) and 51361 (due to Wroblewski).
The comments then list a number greater than 220,000 and that number is 1347505009. The reason I came across these numbers is that today my diurnal age is 27455 and that is the fifth multiple of 5491.
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