## Never square

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• 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 ab-1=u^2 ac-1=v^2
Message 1 of 2 , Oct 3, 2002
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

ab-1=u^2
ac-1=v^2
bc-1=x^2
bd-1=y^2
cd-1=z^2

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.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths
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• ... 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
Message 2 of 2 , Oct 3, 2002
--- 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

Note that (3*a+2) is equal to -1/4 (modulo 4*a+3).
[It is the modular solution to 4x+1 == 0]