Recent content by lifom

Many thanks. I have found another proof on the web. May I ask the generalization: Let S(m) be the number of ways to express m = a^2+b^2, a>=b. S(p) = 1 if p = prime = 1 mod 4 (proved) S(p) = 0 if p = prime = 3 mod 4 (proved) S(pq) = 2 if p,q are distinct prime = 1mod 4 (proved)...

Thanks, But I haven't learned ring yet. So would you mind to suggest a eaiser proof?

I know that any prime p = 1 mod 4 can be expressed as sum of 2 squares. But how many different pairs of integers a,b such that p = a^2+b^2? (with a>b!) It seems there is only one pair. How to prove it? I try in this way: assume p = a^2+b^2 = c^2+d^2 (with a>c>d>b) and try to show it has...

Let m be a product of distinct primes p1,p2,...pr. Assume x=c (mod pi) is a solution of x^k=b (mod pi) for all i =1,2,3,...r. Can I conclude that x=c (mod m) is a solution of x^k=b (mod m) ? (I think that if a=b mod x and a=b mod y then a=b mod(xy), provided that gcd(x,y)=1)

I get it! Thanks.

I have tried, but fail in the last step: In Wilson's proof, the number satisfy a^2=1(mod p) -> (a-1)(a+1) = p*n. Since 0<=a<p a= 1 or p-1 only. So for 2,3,...,p-2, I can group them into pairs without repeat such that the pair product = 1 (mod p)... But in my problem, the number satisfy...

Let b1,b2,...bn be the integers between 1 and m that are relative prime to m (including 1), and let B = b1*b2*...*bn be their product. The quantity B came up during the proof of euler formula. a^n = 1 (mod m), where n is number of integers between 1 and m that relative prime to m. How can I...

I think it is not true. Example: the largest number not in the form 3x+7y is 3*7-3-7=11. That means it can be 12,13,14,... and so on. So any integer > 11 can be in the form of 3x+7y. Do you mean there are exactly only (a-1)*(b-1)/2 positive integers NOT with the form ax+by where...

(p-1)! = -1(mod p), where p is a prime I have tried small values of p but I can't find any pattern. Can anyone give me some hints or directions? I don't know a detail proof. Thank you

Thanks Dodo. I have written the proof. Let P(n) : "n is a form of ax+by, gcd(a,b)=1, a,b are positive integers, x,y are non negative integers" Clam 1: If P(n1) and P(n2) is true, then P(n1+n2) is also true. (since n1=ax1+by1, n2=ax2+by2, then (n1+n2)=a(x1+x2)+b(y1+y2) x1+x2>=0,y1+y2>=0)...

Consider 3x+7y, (x,y are non negative number),gcd(3,7) =1. some nonegative integers cannot be in form of 3x+7y (example, 3x+7y can't be 1,2,... the largest number not in the form of 3x+7y is 11. (3*7-3-7) so I guess the largest number not in the form of ax+by,gcd(a,b)=1 is ab-a-b. But I don't...

I have tried some values of a,b. And I guess the largest number not in the form of ax+by, (x,y are non-negative integers) with gcd(a,b)=1 is ab-a-b. But I don't know how to prove it. Can anyone give me some hints or directions? I don't want the detail proof,as I want to try it by myself. thank you

I use computer to compute from 1 to 1000, I still can't find 3 different ppts with same values of c. Can you give me an example?