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Is it true that every positive integer is the sum of $18$ fourth powers of integers?

Does this means every positive integer $n = x_1^4+x_2^4+\cdots+x_n^4$ for some positive integer $n=18$? Could you show me some example, I don't think I am understanding the question..( ex: Explanation says $78$ can be written as a sum of $18$ fourth powers of integers, how? Can $1$ or $5$ be written as sum of $18$ fourth powers of integers?)

Thanks!

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    $\begingroup$ using underscores makes subscripts more readable; e.g. x_1^4 gives $x_1^4$. $\endgroup$ – robjohn Aug 1 '11 at 18:16
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    $\begingroup$ 0 is a fourth power. $\endgroup$ – lhf Aug 1 '11 at 18:19
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    $\begingroup$ 1 can be written as $1^4$, and 5 can be written as $1^4+1^4+1^4+1^4+1^4$ The trick is then for numbers larger than 18, where is basic trick does not work. Think of this trick: given a number n, look at the largest power $x_1^4$ that is smaller than n, then look at the difference n-$x^4$, and find the largest integer $x_2$ with $x_2^4$< n-$x^4$, and so on. $\endgroup$ – gary Aug 1 '11 at 18:21
  • $\begingroup$ @gary: Your "greedy" algorithm is not the best! For instance, it decomposes $160$ into $20$ fourth powers $(81+4 \times 16+15 \times 1)$, but we know that $19$ fourth powers are always enough $(160=10 \times 16)$. $\endgroup$ – TonyK Aug 2 '11 at 9:57
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This is Waring's problem for $k=4$. The answer to your question is no because 79 requires 19 fourth-powers, as reported in the wikipedia page.

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Note that some of the integers may be 0. So $1 = 1^4 + 0^4 + \ldots + 0^4$. $78 = 4 \times 2^4 + 14 \times 1^4$. But what about 79?

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This is called Waring's problem and is a nontrivial problem. It is actually known that there are numbers which cannot be written as the sum of $18$ fourth powers. Waring's problem states that every number, however, can be written as the sum of $19$ fourth powers.

As for your example, $1$ is the sum of $18$ cubes since

$$ 1 = 1^4 + \underbrace{0^4 + \dots 0^4}_{17~\text{times}}. $$

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    $\begingroup$ Don't you mean fourth powers instead of cubes? $\endgroup$ – lhf Aug 1 '11 at 18:17
  • $\begingroup$ @lhf: Of course. Thank you for the correction. If you feel my answer is too close to yours, I'm happy to erase it. $\endgroup$ – JavaMan Aug 1 '11 at 18:20

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