CollegeAnon !LAbIRp9cT. ID: b6f6bb April 5, 2018, 10:21 a.m. No.5453   🗄️.is 🔗kun

>>5452

So we would want our e to be a negative square. Interesting because negative squares have entries in every column and they are easy to generate (i think).

 

If d^2-r^2 = c then d^2 + e = c where e = -r^2 would be a solution. This would imply that (d-r)(d+r) = c which means d-r = a, d+r = b.

 

If d-r=a then x = d-a = d-(d-r) = r

Then, since (d+n)^2 - (x+n)^2 = c and we already know that x = r and d^2 - x^2 = c so n must be zero. If n is zero then xx+e must be zero also.