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I think as the number increases , prime become more rare , it'll be easier to do it.
I 've though of sth , but don't know whether it's useful or not , but can speed up a little
For example , the number 679543
if
we just need to find p.
But the range is still to large , then assume 679543=qpm , then
Assume that 679543=abcd , then
Try , 23,19,17,13,11, 7 5, 3, 2, ,find out that none of them can divide the number , then the number may be the product 3 primes or 2 primes .
I don't know if this can help.
Can be done on Maple
Nice Technique , George.~
Yah~true , that will bring a huge revolution ~hehe and turbulence
it's just my opinion , if you see w as a constant , then the derivative will be
where w equal to 1301) His weight remain constant , you can see w as a constant then find the derivative
but where is point o?
I havent really learned about limit, calculus and stuffs , never take them seriously, When I saw this one, I thought of
, too .We can divide n by a the prime we find because we know that n is a product of distinct primes, and so once we find one, we can "take it out" of n.
but if a number is the product of two 22 digits primes , then ....
If I can find a method ~WOW~lol ~jk
Don't let this one sink , It's a good question ,guys. I can't work it out tho
Lol~ that's programming method ,right?
The generator is carried out by observation of X^2+Y^2=Z^2 , if you put them back into the original formula , you will find , two side of it are equal .
but there's little beyond my knowledge ,
Oh , I miss your proof , sorri~ Yep , that's it
then I shall prove
I shall show 5 will always divide one of the following: Hmmm,Shall I consider each set of ridues ?According to above , the two numbers of yours , (A,B)=1 , then there must be p,q such that Ap+Bq=gcd(A,B)=1
Here is another version of proof from my textbook
Let H be the subset of Z consisting of all integer of the form
The unique factorization theorem says that every positive integer number greater than 1 can be written in only one way as a product of prime numbers.
Is there a way I can find those prime faster? so I don't have to try all the primes less than that integer.
Right~Exactly
As in my early post , the generator of the triplet
, m, n random integers.What you mean is
right? if so , it's a parabolaOh ,yeah , If I just prove
for there is always an odd k that makes it be divided by 5 , will that work?