>>5564

VQC I was thinking could we do something with the equation x^2 + e as a function of x. Also I learned in Number Theory about quadratic forms, which are this:

ax^2 + bxy + cy^2 and you can denote them like this (a,b,c). This can be used to represent integers

(3,1,2) would be equivalent to (2,1,3).

In addition, another odd equivalence is (2,0,3) and (2,4,5). A way to generate these equivalent forms is through a unimodular substitution such as x = pX+qY and y=rX+sY. Even though these may generate different equations they are equivalent statements. But due to some stuff, you need ps-rq =+/- 1. Therefore different forms can be used to represent the same integers.

Def: Integer is properly represented by (a,b,c) if n=ax^2 + bxy + cy^2 where gcd(x,y)=1.

Def: The Discriminant d = b^2-4ac (also what I suspect d to be). If two forms are equivalent they have the same determinant (doesn't go the other way). If two forms have the same discriminant but aren't equivalent then they are disjoint. Also, discriminants must be congruent to 0 or 1 (mod 4).

If d<0 it is called definite. If definite, a>0 represents positive integers, a<0 represents negative integers. d>0 is definite, sometimes positive sometimes negative.

Given a form (a,b,c) we want to know if we can represent n by the form, where gcd(x,y) = 1.

Method: Supposes (a,b,c) represents

n = ap^2 + bpr + cr^2 gcd(p,r)=1

Want to find q,s such that ps-qr = 1 (easy to do euclids algorithm for this)

Then substitute x=pX+qY, y=rX+sY

then, you have

AX^2 + BXY + CY^2 = nX^2 + BXY + CY^2

also -A = ap^2 + bpr + cr^2 = n

So, if (a,b,c) properly represents n, then there is a form (n,h,L) (h,L not known) that also is equivalent. Also, if (a,b,c) represents n properly, then since discriminants are the same you know b^2-4ac = h^2-4nL. Also h^2 is congruent to d (mod 4n ) [n is absolute value if it was negative]. If we pick h, then L is fixed. We can test the congruence to see if n is or is not represented by this. Also, due to the congruence, we can assume that 0<h<2n. Watch this now. This is where I think it connects to the grid.

nx^2 + hxy + Ly^2

x = X + Y , y = Y

=n(X+Y)^2 + h(X+Y)(Y) + L(Y)^2

=nX^2 + 2nXY + hXY + hY^2 + lY^2

=nX^2 + XY(h+2n) + (h+L)Y^2

Now look at the middle h term, which we were picking. We can just add 2n to this to generate a new form. Teacher said to shift it until 0 < h < 2n. Remember, if we add 2n to the middle term, then since we have the same discriminant the next L would be different, but its easily calculable.

So to try and connect it here, the equation we would be solving would be (1,0,-1) for x=d+n, y=x+n because that would be a difference of two squares. Moreover, it could be (0,1,0) for x = a, y = b. My thinking is that since we can shift by 2n in this equation, then if we are aligning it with the grid, the middle term h would be the e value for (e,n,d,x,a,b) because that is the one you would be shifting by 2n.

>>5562

>>5567

a is odd, so if x is even, then d is odd. If x is odd, then d is even.

(x+n) = 2j+1

(n-1) = 2j-x

Suppose x is even, this means that d is odd and n is odd also. Then, (d+n) can take values 2k so

2k = (d+n) = (d+ 2j-x + 1)

= a + 2j-1

= a + (x+n)-2

(d+n) = 2(k-1) = a + (x+n)

Suppose x is odd, then d is even, and n is even. Then, (d+n) takes values 2k also actually the same thing, but I think this is valuable

(d+n) = 2(k-1) = a + (x+n)

>>5562

Attempt like 7 on this

Suppose

x = 2j+1,

(x+n) odd -n even

n = 2k

n-1 = 2k-1

x+n = 2(j+k) + 1

(d+n) even and n even =d even

d = 2p

f = 2d+1 = 2(2p) + 1

f = 4p+1

d - x = a -2p - 2j - 1 = a

a = 2(p-j) - 1

b = a + 2x + 2n

b = 2(p-j) - 1 + 2(2j+1) + 2(2k)

b = 2p - 2j - 1 + 4j + 2 + 4k

b = 4k + 2p + 2j + 1

Suppose x = 2j

(x+n) odd -n odd

n = 2k+1

n-1 = 2k

(x+n) = 2(j+k) + 1

(d+n) even n odd =d odd

d = 2p+1

f = 2(2p+1) + 1

f = 4p+3

a = d-x = 2p+1 - 2j = 2(p-j) + 1

b = a + 2x + 2n = 2(p-j) + 1 + 2(2j) + 2(2k+1)

b = 2p + 2j + 4k + 3