From (-f,n-1) = c, the value of a,b and d increase by ONE every 2(n-1) cells, as you move from left to right in the grid.

From (e,n) = c, the value of a,b and d decrease by ONE every 2n cells, as you move from right to left in the grid.

These two rows are next to each other.

Any product of 2 primes will be divisible by 3 if you add either 2 or 4, or if you subtract 2 or 4.

Any product of 2 primes will be divisible by 5 if you…

Etc…

Remainders.

Patterns.

"Triangulation"

>>6385

Almost.

Yes four orange squares, values of n will be different.

Well done.

The values of n at these cells can be calculated.

>>6378

p and q were just distinguishing from specific e and n for that example.

Just integers.

Signed integers is a hint at using assymetry of -f and e where possible.

>>6380

Thank you.

>>6400

>>6400

Spoopy but coincidence from my end.

Think about the two rows.

Each c has patterns of remainders.

Those two rows next to each other give away a LOT of information.

They cover a different set of gaps. The bottom one is a unit of two more.

The difference in increase 2(n-1) and decrease 2n give enough information, and make the product of two primes work to your advantage!!!

Triangulate.

You can do this.

Given two columns f+e or 2d+1 apart.

-f

e

How can we tell or construct the list of factors in a[t] that are 1 unit apart?

Those factors represent the cells with elements in each column.

The values of a[t] in each column are all the possible values of x squared plus e (or -f).

The key to using the grid is at hand.

The difference in the element position of b(n-1) and bn by ONE element at (-f,1) and (e,1).

What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.

>>6442

This.

The values of a,b and d each increase by ONE 2n cells to the right. They decrease by one 2n cells to the left.

For every c.