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My question is actual at the bottom in bold and is about the logic the author uses.

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

This is a question form a discrete math book, not to worry the answer is detailed in the book as well:

Solution: To determine whether a positive integer n can be written as the sum of 18 fourth powers of integers, we might begin by examining whether n is the sum of 18 fourth powers of integers for the smallest positive integers. Because the fourth powers of integers are 0, 1, 16, 81, . . . , if we can select 18 terms from these numbers that add up to n, then n is the sum of 18 fourth powers. We can show that all positive integers up to 78 can be written as the sum of 18 fourth powers. (The details are left to the reader.) However, if we decided this was enough checking, we would come to the wrong conclusion. It is not true that every positive integer is the sum of 18 fourth powers because 79 is not the sum of 18 fourth powers (as the reader can verify).

my question is about the authors logic. He specifically says that the fourth powers are 0, 1, 16, 81. i presume because 0^4, 1^4, 2^4...

so my reasoning says that i'm to select 18 terms [out of a set [0,1,16,81...] and i should be able to create any number with them

I choose to try to create the number 3

so [0,1,16... eh]

well 1 is less then 3, but after that i can't draw anymore because i'll go over 3. so the proof is false (which the author confirms)

BUT the author says " can show that all positive integers up to 78 can be written as the sum of 18 fourth powers. (The details are left to the reader.)" apparently he shouldn't have left them up to me because I'm not getting it. So how did he say produce 3 out of 18 fourth powers? or more likely what aren't i grasping here in this example

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    $\begingroup$ Note that $$79 = 64 + 15 = 16 + 16 + 16 + 16 + 1 + \cdots +1 = 2^4 + 2^4 + 2^4 + 2^4 + 1^4 + 1^4 + \cdots + 1^4$$ requires 19 fourth powers. I'm just saying. $\endgroup$
    – Will Jagy
    May 29, 2012 at 23:30

3 Answers 3

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What you’ve not realized is that repetitions are allowed:

$$3=\underbrace{0^4+\ldots+0^4}_{15\text{ copies}}+1^4+1^4+1^4\;.$$

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You're allowed to use a power more than once in this context. So in this case $3 = 1+1+1+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0$.

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We, can basically repeat the term, so the terms which we can use are 0, 1 ,16. Now, maximum is 18 terms: 16 + 16 + 16 + 16 = 64, we are now left with 14 terms and we can use 14 1's for that, so total terms will be 18. As 16 is the maximum we can use for creating in 18 terms and in creating 78 we have used maximum 16 possible, so 79 won't be possible with 18 terms. We, can do this proof by all cases: $$16K1 + 1K2 + 0K3 \le 78$$ $$K1 + K2 + K3 = 18 \text { and } 0 \le K1 \le 4$$ So, we can check by taking $K1=0, 1, 2, 3, 4$ and all terms till $78$ are covered

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  • $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – Integrand
    Sep 13, 2020 at 13:57
  • $\begingroup$ The author did know about repeating numbers was allowed. So, in my answer I have stated that number repeating is allowed. $\endgroup$ Sep 13, 2020 at 15:15

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