Given two integers a and b such that:
ii. a+b divides ab+1
iii. a-b divides ab-1
Prove that a is less than b*sqrt(3).
(In reply to re: Thoughts...hints toward formal proof
by Ady TZIDON)
Ady's hints imply that there are two lemmas to prove in order to complete the main proof.
Lemma 1: Prove b^2-1 must be divisible by a+b and a-b.
Lemma 2: Prove the greatest common divisor of a+b and a-b is either 1 or 2.
Lemma 1 proof:
Start with the identity b^2-1 = b^2+ab-ab-1.
Rewrite the identity as b*(a+b) - (ab+1). The first term is divisible by a+b as implied by its factoring. From statement ii the second term is divisible by a+b. Therefore the entire expression is divisible by a+b.
Now rewrite the identity as -b*(a-b) + (ab-1). The first term is divisible by a-b as implied by its factoring. From statement iii the second term is divisible by a-b. Therefore the entire expression is divisible by a-b.
The combination of these two conclusions then implies b^2-1 must be divisible by both a+b and a-b. QED Lemma 1.
Note that Lemma 1 does not imply b^2-1 is divisible by the product (a+b)*(a-b), for example a=19,b=11 yields the fraction (121-1)/(30*8) = 120/240 = 1/2. This is because the greatest common factor of a+b and a-b is doubly represented in the denominator but only singly represented in the numerator. Now Lemma 2 comes into play.
Lemma 2 proof:
Let g be the greatest common divisor of a and b. Then g is a factor of a+b and a*b. a*b is coprime to a*b+1. g is a factor of a*b therefore g is coprime to a*b+1.
a*b+1 is divisible by a+b and a+b has g as a factor, therefore g is a factor of a*b+1. However g is coprime to a*b+1. Then g must equal 1, which immediately implies a and b are coprime.
Let h be the greatest common divisor of a+b and a-b. h must then also be a factor of the sum and difference of a+b and a-b. (a+b)+(a-b) = 2a and (a+b)-(a-b) = 2b. Then h is a common factor of 2a and 2b.
With a and b being coprime, the only possible values of h are the factors of 2, namely 1 and 2. Therefore the greatest common divisor of a+b and a-b is either 1 or 2. QED Lemma 2.
With both lemmas proven we can now state 2*(b^2-1) is divisible by the product (a+b)*(a-b).
Any positive integer is greater than or equal to any of its factors. Then 2*(b^2-1) >= (a*b)*(a-b). Rearrange this inequality into 3*b^2 - 2 >= a^2. Adding 2 to the left side will make the inequality strict: 3b^2 > a^2.
From statement i, a and b are both positive. This means we can take the square root of each side without disturbing the inequality. Doing so yields sqrt(3)*b > a. QED