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Is $524154366113525716400386$ the sum of two fourth-powers? I suspect that this number is the sum of two fourth powers. Can anyone use a computer program, or SAGE, or Wolfram Alpha to check whether this number is the sum of two fourth- powers?

The complete factorization of this number is given by : $$524154 366113 525716 400386 = 2 × 521 × 8761 × 21529 × 221281 × 12 052297$$ Notice that all the factors are of the form $16n+1$ or $16n+9$ (They are all congruent to $1$ or $9 \mod16$ ). Hence this number may be the sum of two-fourth powers.

Mythomorphic
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Derak
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    See also this question. – Dietrich Burde Aug 18 '17 at 11:48
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    where are these numbers coming from? – supinf Aug 18 '17 at 11:51
  • If it is such a sum, then one of the powers is divisible by $5$, since the final digit of a 4th power is 0, 1, 5, or 6, and your number ends with $6$, hence comes from either $ 1 + 5$ or $0 + 6$. – John Hughes Aug 18 '17 at 11:51
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    I used PARI/GP and brute force : The number isn't the sum of two $4$-th powers. – Peter Aug 18 '17 at 11:53
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    @Peter . I am searching for counterexamples to the Euler Quartic Conjecture . This particular number turns up in my search and as far as I can tell, it has all the right properties to make it the sum of two fourth powers. For example , this number is congruent to 9 (mod 13) !! – Derak Aug 18 '17 at 11:59
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    @Derek: afaik the euler quartic conjecture is already disproven – supinf Aug 18 '17 at 12:01
  • ? n=524154366113525716400386;k=sqrtnint(n,4);for(a=0,k,if(ispower(n-a^4)>0,print (a))) is the program I used. There is a more efficient way (necessary for larger $n$), but I do not remember at the moment how it works – Peter Aug 18 '17 at 12:03
  • @supinf I guess, Derek is searching for more counterexamples. – Peter Aug 18 '17 at 12:04
  • @Derek It was a difficult task to disprove the conjecture you mentioned. Numbers do not tend to be the sum of two $4$ th powers, a sum of two squares occurs significantly more often. – Peter Aug 18 '17 at 12:06
  • @Peter a sum of two fourth powers is a sum of squares of squares I believe to be the cause. –  Aug 18 '17 at 12:08
  • @RoddyMacPhee Perhaps, I formulated it unlucky. The chance that a number is the sum of two $4$-th powers is much less than the chance that it is the sum of two squares. – Peter Aug 18 '17 at 12:12
  • I get that but to be a sum of fourth powers it is a necessary condition that it be a sum of two squares (of squares). –  Aug 18 '17 at 12:21
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    @JohnHughes: Both last digits need to be odd because there is only one factor of 2 in the final sum; therefore 0+6 can also be eliminated as well. $(a+b)^4+(a-b)^4=2\left( \left( a^2+b^2 \right)^2 +(2ab)^2 \right)$ – James Arathoon Aug 18 '17 at 12:37
  • @Peter oh and another thing in this case sqrtint(sqrtint(n)) is actually faster. –  Aug 18 '17 at 14:52
  • @RoddyMacPhee sqrtnint(n,4) is the truncated value of $n^{1/4}$ – Peter Aug 18 '17 at 21:36
  • I know I play around in PARI/GP console a lot. sqrtint(sqrtint(n)) is smaller ( okay tested it maybe not but it was slightly quicker to use) yet can be proven to be an upper bound. fourth powers are squares of squares. So an integer square, that doesn't exceed n, needs to have a base of at most sqrtint(n). but we also need the base to be an integer square. The largest integer value that can work as the base for this square is sqrtint(sqrtint(n)) hence it's the upper bound on any value whose fourth power doesn't exceed n, as well as the other properties needed. –  Aug 18 '17 at 21:51

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In case you were not aware, Mathworld mentions Elkies' disproof, citing On $A^4+B^4+C^4=D^4$.

Without having any software installed on your computer, you can check that this particular number is not the sum of two fourth powers by, say, pasting FindInstance[(524154366113525716400386524154366113525716400386==x^2+y^2)~An‌​d~(x==z^2)~And~(y==w‌​^2),{x,y,z,w},Intege‌​rs] into the wolfram cloud sandbox and use Shift+Enter or numpad Enter or clicking on Gear>"Evaluate Cell" to run it. (For confirmation that this code would work, compare to using 280286069726155265499093303843106.)

There may be some modular congruences that rule this number out as well.

Mark S.
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