Interesting way to define a sequence of Squares

[ramsey2879]

Let A and B be two coprime integers. Find X = zero mod A such that Y = 2*X +1 = 0 mod B. Then 8*(Y +2*N*A*B)*(X + N*A*B) + 1 is a square for all integer N.

If A = 5,B = 7, X = 10, Z = 21 then the sequence of square roots of the Squares for N = -3 to 3 is -379, -249, -99, 41, 181, and 321. Of course X is also variable since X could also equal 45 then Y would be 91.

 

[atomthick]

sorry, what is the question?

 

[ramsey2879]

sorry, what is the question?

It would be great if someone could prove or help me prove that my conjecture is valid for all coprime A and B. So far I have only specific examples to show. I also believe that the square roots are = 1 mod A and = -1 mod B if you use a positive or negative square root respectively for n >= 0 or < 0, so that might be a start to making a proof.

 

[dodo]

Hi, ramsey,
the example you give, A=5, B=7, X=10, Y=21, with N=1 gives 35841, which is not a square. In fact i haven't yet obtained any square with the first few numbers I plugged in. Which numbers did you use to obtain squares?

 

[ramsey2879]

Hi, ramsey,
the example you give, A=5, B=7, X=10, Y=21, with N=1 gives 35841, which is not a square. In fact i haven't yet obtained any square with the first few numbers I plugged in. Which numbers did you use to obtain squares?

Sorry The equation should be 8*(Y+2*N*A*B)*(X+N*A*B)+1 = S^2 I misplaced the 2 in my first post. I think I have it right now. Got to go.

 

[dodo]

Ah, in this case 8*(2*X+1 + 2*N*A*B)*(X + N*A*B) + 1 simplifies to the square of 4*(X + N*A*B) + 1; in other words, it is a square for any choice of A,B,X,N, always, without any conditions about coprimality or congruency.

http://www.wolframalpha.com/input/?i=factor+8*(2*X+1+++2*N*A*B)*(X+++N*A*B)+++1

 

[ramsey2879]

Ah, in this case 8*(2*X+1 + 2*N*A*B)*(X + N*A*B) + 1 simplifies to the square of 4*(X + N*A*B) + 1; in other words, it is a square for any choice of A,B,X,N, always, without any conditions about coprimality or congruency.

http://www.wolframalpha.com/input/?i=factor+8*(2*X+1+++2*N*A*B)*(X+++N*A*B)+++1

Good Now I can say that I have a proof that if C*D = X*(2*X +1) and A = GCD(C,X) and B = GCD(D,2X+1) then 8*(C+2*N*A^2)*(D+N*B^2)+1 = S^2 for all integer N because I got from that conjecture to the result of my post. See if you can see the connection between the two!

 

[ramsey2879]

Good Now I can say that I have a proof that if C*D = X*(2*X +1) and A = GCD(C,X) and B = GCD(D,2X+1) then 8*(C+2*N*A^2)*(D+N*B^2)+1 = S^2 for all integer N because I got from that conjecture to the result of my post. See if you can see the connection between the two!

What is interesting about this is that the pairs (C,D) and (E,F) where E = C+2*A^2 and F = D + B^2 each form the starting terms of two recursive series of the form S(n) = 6*S(n-1) - S(n-2) + K where K is dependent upon the starting terms, i.e. K/2 = Square root of (8*C*D+1) -C - D and likewise substituting E,F for C,D. and S2(2n)-S1(2n) always equals twice a square and S2(2n+1)-S1(2n+1) always equals a square.

An example is S1 = {0,1,6,35,204,1189 ...} S2 = {0,2,14,84,492,2870 ...} The differences are {0,1,8,49,288,1681,...). Because of the previous posts, for any Coprime pair A,B can form the series 2*A^2, B^2, 2*C^2, D^2, 2*E^2, F^2 ... having the recursion S(n) = 6*S(n-1) - S(n-2) + T where T is dependent upon the pair A,B. Of course the product of two adjacent terms in either series S1, S2 is always a triangular number. I can give a proof of this if anyone is interested.

 

[dodo]

I'd be interested in a proof of the statement

if C*D = X*(2*X +1) and A = GCD(C,X) and B = GCD(D,2X+1) then 8*(C+2*N*A^2)*(D+N*B^2)+1 = S^2 for all integer N

On the part about the series, frankly speaking, I have no particular interest. But I'm interested in how you used the GCD conditions in the proof of the statement above.

 

[ramsey2879]

I'd be interested in a proof of the statement


On the part about the series, frankly speaking, I have no particular interest. But I'm interested in how you used the GCD conditions in the proof of the statement above.

If C*D = N*(2*N+1) and GCD(C,N) = A and GCD(D,2*N +1) = B you can prove that 8*(C + 2*n*A^2)*(D + n*B^2) +1 = S^2 as follows:
It can be shown that C = A*(2*N + 1)/B and D = B*N/A. Making these substitutions you get the following:
S^2 = 8*(A*(2*N+1)/B + 2*n*A^2)*(B*N/A+n*B^2) + 1
S^2 = 8*(2*N+1)/B + 2*n*A)*(B*N + n*A*B^2) + 1 :Repositioning the factor A
S^2 = 8*(2*N +1 + 2*n*A*B)*(N + n*A*B) + 1 :Repositioning the factor B
S^2 = 8*(2*X+1)*X + 1 = 16*X^2 + 8*X + 1 : X = N + n*A*B
S^2 = (4*X + 1)^2

Is this what you wanted? I don't think you need more explanation but I can explain further if you want.

 

[dodo]

It can be shown that C = A*(2*N + 1)/B and D = B*N/A.

Ah, thanks, that was the part I found interesting. If CD = N(2N+1), or C/N = (2N+1)/D, then, upon reduction of these two fractions (which involves dividing up and down by A=GCD(C,N) for the left fraction, and by B=GCD(D,2N+1) for the fraction on the right), after reduction, the numerators will be equal and the denominators will also be equal, so C/A = (2N+1)/B and N/A = D/B, from which the result follows.

Thanks again for the hint!