I Want Four

Number Theory Level 5

Lagranges four square theorem states that every positive integer can be written as the sum of four integer squares, possibly zero.

How many odd integers cannot be represented as the sum of four non-zero squares?

The answer is 8.

1 solution

Michael Mendrin
Jan 24, 2017

It's not hard to try out combinations of $(1, 4, 9, 16, 25, 36, 49, 64, 81,100)$ to see that the only odd integers that fall through are $(1, 3, 5, 9, 11, 17, 29, 41)$ , and then that's it. But proving that it ends with $41$ is quite another matter.

Yes, that's right, I guessed it would end at $41$ . Math intuition.

To be fair, I don't have a great reason why it ends at 41.

If I had to write a complete solution, it would be tedious checking building up iteratively from "sum of 2 non-zero squares" and "sum of 3 non-zero squares".

Calvin Lin Staff - 4 years, 4 months ago

You'd want to think this would be something like Frobenius Number , showing $41$ top be the "largest number with no solution". But it don't apply here. Close but no cigar.

I'll think on it.

Michael Mendrin - 4 years, 4 months ago

FYI There are infinitely many even numbers which cannot be written as the sum of 4 non-zero squares, like $2 \times 4 ^ k$ .

Calvin Lin Staff - 4 years, 4 months ago

@Calvin Lin – That's right, so much for any "Frobenius Number" kind of proof. This is intriguing but is not going to be easy.

Michael Mendrin - 4 years, 4 months ago

I Want Four

The answer is 8.

1 solution

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