- Anyone know why 4ab+3(a+b)+2 is never a square?
It comes from looking at the 'alternative' 5-squares conjecture, find a set
a,b,c,d such that
I think no such 4-tuple exists.
Anyway...(4f+3)(4g+3)-1 = 16fg+12(f+g)+8 = 4[4fg+3(f+g)+2]
(4f+3)(4g+1) is obviously never a square.
- --- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> Anyone know why 4ab+3(a+b)+2 is never a square?Rewrite this as:
N = (4*a+3) * b + (3*a+2)
In order for N to be a square, we must have (3*a+2) be a
quadratic residue modulo (4*a+3).
Note that (3*a+2) is equal to -1/4 (modulo 4*a+3).
[It is the modular solution to 4x+1 == 0]
-1/4 is a quadratic residue iff -1 is a quadratic residue.
And finally, -1 is never a quadratic residue modulo 4*a+3;
this is a basic result of quadratic reciprocity.
Or for you PARI/GP types who don't care about all of the theory:
...prints out an endless stream of -1, which gives strong
evidence that 3*a+2 can never be a quadratic residue