Proving the Unsolvable: Lagrange's Theorem and 4 Squares

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SUMMARY

The discussion centers on proving that at least one of any four consecutive natural numbers cannot be expressed as the sum of two squares, as stated in Lagrange's Theorem. Participants analyze the problem using modular arithmetic, specifically considering the properties of squares modulo 4. It is established that the square of any natural number modulo 4 results in either 0 or 1, leading to the conclusion that among four consecutive numbers, one must be 3, which cannot be represented as a sum of two squares. This contradiction confirms the theorem.

PREREQUISITES
  • Understanding of Lagrange's Theorem in number theory
  • Familiarity with modular arithmetic, particularly modulo 4
  • Basic knowledge of squares and their properties in mathematics
  • Ability to construct mathematical proofs
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  • Study the implications of Lagrange's Theorem on number representations
  • Explore modular arithmetic in greater depth, focusing on applications in number theory
  • Learn about the properties of quadratic residues and their significance
  • Investigate other proofs related to sums of squares, such as Fermat's theorem on sums of two squares
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Mathematicians, students of number theory, and anyone interested in the properties of integers and their representations as sums of squares.

RichardCypher
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Hi everybody :smile:
I'm currently reading Burton's Elementary Number Theory (almost done!) and in the chapter about Lagrange's Theorem about the sum of four squares, there is a supposedly easy question which I can't solve for some reason :blushing:. I'd really appreciate a hint or two...

Prove that at least one of any four consecutive natural numbers is not a sum of two squares [that is, can't be represented as the sum of two squares of whole numbers]

Thank you all! :smile:
 
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Well I think this one works?

2 + 3 + 4 + 5 = 14

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16

It has to be the SUM of TWO squares... 9 + 4 = 13. None of the others work.
 
Consider four consecutive numbers mod 4, then consider squares mod 4. The result follows immediately.

iwin2000: the problem was to show the result for all {n, n + 1, n + 2, n + 3}, not just for one such instance.
 
CRGreathouse said:
Consider four consecutive numbers mod 4, then consider squares mod 4. The result follows immediately.

iwin2000: the problem was to show the result for all {n, n + 1, n + 2, n + 3}, not just for one such instance.

The square of any natural number mod 4 has to be 0 or 1. Therefore, the sum of two such squares mod 4 has to be 0, 1, or 2. However, out of four consecutive natural numbers mod 4, one has to be 3. Contradiction. Is that right?

Great hint! Thank-you very much :biggrin:
 
You got it.

I like minimal hints.
 

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