• ... The flaw was: how do you know that n is /prime/? ... The mend is, interstingly enough, to make the theorem /stronger/. After all, the primeness or
Message 1 of 2 , Jan 29, 2007
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I wrote (about a purported proof):

> However, I now realize (having pasted it here and stared at it
> again), that it has a fatal flaw.

The flaw was: how do you know that n is /prime/?

> maybe someone can mend it?
>

The mend is, interstingly enough, to make the theorem /stronger/.

After all, the primeness or otherwise of a and b plays no strong role.
If the restriction on primality is removed, experiment shows the only
solution for min(a,b) < 16M is the trivial solution (1,1).
The theorem and proof follow.

For rational integer x, let f(x)=x^2-x+1.
Theorem:
The only pair of postive integers (a,b) such that (a*b) | f(a)*f(b)
is (1,1).

Proof:
Suppose the contrary, and let (a,b), a <= b, be the minimal pair.

No prime factor of a divides f(a), so each must divide f(b), so for
some m < b:
m*a = b^2-b+1
No prime factor of b divides f(b), so each must divide f(a), so for
some n < a:
n*b = a^2-a+1

n^2 = n^2 mod a
n^2*b = n mod a
n^2*b^2 = 1 mod a
So
n^2*(b^2-b+1) = n^2-n+1 mod a
So
n^2*(m*a) = n^2-n+1 mod a
So
n^2-n+1 = 0 mod a
i.e. a | f(n)
But n | f(a)
Also, n < a.
So (n,a) is a solution which is smaller than the solution (a,b).