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Can $406014677132263504491682$ be the sum of two fourth powers? It may be so. Can anyone use Wolfram Mathematica , SAGE , or some computer program, to check whether this number is the sum of two fourth powers ?

The complete factorization of this number is given by : $406014 677132 263504 491682 = 2 × 1459 × 6883 × 21529 × 938976 705857$

Some Guy
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Derak
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    Where are these numbers coming from and is there any reason you aren't using wolfram alpha? – abiessu Jul 18 '17 at 21:58
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    ._. The majority of your questions are these seemingly random "Can ____ be written as the sum of ____ ____th powers?" Surely you must've learned something from your previous questions so that you could better show your attempts at the problem? – Simply Beautiful Art Jul 18 '17 at 22:00
  • I am searching for counterexamples to the Euler Quartic Conjecture . These numbers turn up as part of my search for counterexamples . Unfortunately Wolfram Mathematica , and Wolfram Alpha are unable to solve this type of problem. They are not reliable. Sometimes they claim there are no integer solutions, whereas in fact there may be ! – Derak Jul 18 '17 at 22:03
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    Nope, the number can't even written as a sum of two squares because $1459 \equiv 3 \pmod 4$. – achille hui Jul 18 '17 at 22:06
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    If you are looking at this conjecture, presumably you already know of the disproof of it by Elkies (1988) as found on the Mathworld page. So the next step would be to look at that work and figure out what class of solutions was found. Have you done this? – abiessu Jul 18 '17 at 22:10
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    @achillehui gave a simple theoretical answer. Also, in PARI/GP, the simple program a=406014677132263504491682;for(i=0,a^(1/4),ispower(a-i^4,4)&&return(i)) proves by brute force that no solution exists. – Jeppe Stig Nielsen Jul 18 '17 at 22:11
  • @JeppeStigNielsen: I don't program in PARI/GP, but doesn't your code merely test whether $a$ is itself expressible as a fourth power, not that it is expressible as the sum of two terms, each of which is a fourth power? – David G. Stork Jul 18 '17 at 23:57
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    @DavidG.Stork The code looks correct to me, assuming there is a solution, one of the summands would be less than the other and in particular, less than the 1/4th power of a (or 4th power root), call this 'i'. Now go through all possible 'i' and check if (a-i^4) is a fourth power. That seems like what the code does – mdave16 Jul 19 '17 at 00:18
  • @mdave16: Ah... Thanks. Got it! – David G. Stork Jul 19 '17 at 00:19
  • @Derek Yes, Wolfram Mathematica can answer a question like this: For example, type FindInstance[(280286069726155265499093303843106==x^2+y^2)~And~(x==z^2)~And~(y==w^2),{x,y,z,w},Integers] into the wolfram cloud sandbox and use Shift+Enter or numpad Enter or clicking on "Evaluate Cell" to run it. – Mark S. Aug 18 '17 at 12:02

2 Answers2

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You may want to try some program like this in future for testing.

However, I cannot say anything myself about how the program handles precision, etc, with taking $4$th powers and roots of large numbers, but at least you can be sure it's possible if the program returns with a positive (as you can verify it yourself).

https://js.do/

<script>

var a = 406014677132263504491682;

var b = Math.pow(a / 2.0, 0.25);

var c = -1;

for (var i = 1; i <= b; i++) {

    if (Math.pow(a - Math.pow(i, 4), 0.25) % 1 == 0) {c=i; break;}
}

document.write(c);

</script>

Some quick testing with random examples such as $32$ and $280286069726155265499093303843106=12371235^4+129387197^4$ seem to work, though.

Shuri2060
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Hmmmm. $1459^1$. Any whole number where a $4n-1$ prime has an odd exponent can't be rendered as a sum of squares. Let alone fourth powers.

Oscar Lanzi
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