# Is it true that every positive integer is the sum of 18 fourth powers of integers? (No, but i don't understand given answer)

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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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. – Will Jagy May 29 '12 at 23:30

$$3=\underbrace{0^4+\ldots+0^4}_{15\text{ copies}}+1^4+1^4+1^4\;.$$
You're allowed to use a power more than once in this context. So in this case $3 = 1+1+1+0+0+0+0+0+0+0+0+0+0+0+0+0+0+0$.