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I'm just wondering if anyone knows whether or not (1+4k) can be written as the sum of squares or not, and if so, what the proof of that is? Anyone care to share?? Thank you, Carrie Anne Johnson -- What? Because I kill indiscriminately? -Tom Servo
Carrie A. Johnson wrote: > I'm just wondering if anyone knows whether or not (1+4k) can be >written as the sum of squares or not, and if so, what the proof >of that is? Hm... interesting. There is a related problem about every integer being represented as the sum of four squares, but you ask if (1+4k) can be written as a sum of squares, without mentioning a limit on the number of squares. If this is the case, then each number of the form (1+4k) is easily represented as the sum of squares: 4 is represented as 2^2 up to k times, and 1 is just 1^2. So for example 21 is 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2. Pretty cheesy, eh? ;) -- Karl L. Barrus: klbarrus@owlnet.rice.edu keyID: 5AD633 hash: D1 59 9D 48 72 E9 19 D5 3D F3 93 7E 81 B5 CC 32 "One man's mnemonic is another man's cryptography" - my compilers prof discussing file naming in public directories
> I'm just wondering if anyone knows whether or not (1+4k) can be >written as the sum of squares or not, and if so, what the proof >of that is? [primes, that is] There's a nice proof in Chapter 15 of Hardy & Wright. (Need I say the title? _An Introduction to the Theory of Numbers_, still one of the best introductory number theory books around.) The basic reason is that -1 is always a quadratic residue for a prime 1 mod 4. (You can simply calculate this with quadratic reciprocity.) Therefore \exists x: p | ( x^2 + 1 ). This yields an existence after looking at primes in the ring Z[i], the Gaussian integers. If you really want to know more, go buy a copy of the book. It's well worth it. Eric