Can anyone help me with a technical mathematical divisibility issue?

I am pretty mathematical and I don’t understand this question.

Can you express #2 above in terms of d? —> d= ( (k^2) -1 ) /r ?

Shouldn’t (r,s)= a pair, like (0,1) ?

I kinda missed the (r,s)=1 bit, too. I would have thought that was either a domain or a point on a plane, neither of which makes sense in this context.

Sorry… by (r,s)=1 I mean that gcf(r,s)=1, that is, r and s have no factors in common.

I can give a kind of example to make it clearer: let d=2 and k =5, and r=7. Then

1. r is prime
2. r divides d*k^2–1, though now it reads 7 divides 2(5^2)-1=49, and
3. k<r, since 5<7.

The only problem is that the s above is also 7 (since 7*7=49), and I really want r and s to have no factors in common.

Hope that helps.

wish i could i suck at math

how about r=3, d=10, k=2?

r is prime
k<r
s=13, which has no common factor with r.

r=5, d=6, k=4

s=19

@finkelitis: I can’t help you but I did miss seeing you this evening.

Sorry..I don’t speak mathematics anymore. :(

First a little claim that needs proof but I am pretty sure it is easy to prove. (I think it should be proved with the reciprocity law .)

Claim: For any natural number d there is infinitely many primes p such that (d/p) = 1 where (d/p) is the Legendre symbol.

Now let p be a prime, x an integer such that x^2 = d (mod p). If we multiply both sides by (the inverse of x mod p)^2, say y, we get d y^2 = 1 (mod p). Obviously we can take y to be between 1 and p/2. We have d y^2 -1 = p s for some natural number s.

Now if d y^2 – 1 is not divisible by p^2, take:

r = p & k = y

and we are done.

Now assume dy^2–1 is divisible by p^2. Take r = p and k = p-y. We have:

d k^2 – 1 = d (p-y)^2 – 1 = d y^2 – 1 – 2d y p + d p^2 = 0 – 2dyp + 0 (mod p^2)

But 2dy is not divisible by p so, dk^2 – 1 is not divisible by p^2. Which means if we set

dk^2 – 1 = rs, then s is not divisible by r.

On the other hand k = p-y < p = r.

The proof of the claim goes like this:

Let p be any prime of the form 4dn+1. Then if q is a prime factor of d we have:

(p/q) (q/p) = (-1)^ {(p-1)*(q-1)/4}

(This is where I use reciprocity law.)

Now since p = 1 (mod 4) RHS of the above equality is 1. And since p = 1 (mod q) we have (p/q) = 1.

Which means (q/p) = 1 for any prime factor of d. But since Legendre symbol is multiplicative (d/p) = 1 for any prime p of the form 4dn+1.

Thanks everyone for helping, and sorry I’ve been away from this thread for so long. Thanks especially to @BonusQuestion for the proof that there are infinitely many examples of this. My research has become more productive, though there are still some mysterious things going on. The situation above provides me with some interesting examples (though they’re more interesting when s>k as well). I think I’m starting to see what’s going on more clearly.