Does Vieta's Help In Any Way?

Relevant wiki: Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick

$ab - a - b = 1$

$(a - 1)(b - 1) = ab - a - b + 1$

$(a - 1)(b - 1) = 2$

2 is prime so that...

Case 1

$b - 1 = 1$ $→$ $b = 2$

$a - 1 = 2$ $→$ $a = 3$

Case 2

$b - 1 = 2$ $→$ $b = 3$

$a - 1 = 1$ $→$ $a = 2$

The maximum value of b is 3

$(3)(2) - (3 + 2) = 1$

ab-a-b=1 a(b-1)=1+b a=(b+1)/(b-1) a, b positive integers, so let c=b-1, another (positive) integer. a = (c+2)/c = 1 + 2/c For a to be integer, c must be a factor of 2, so b-1 is a factor of 2. Largest b comes from largest b-1, so b-1 =2 (largest factor of 2). b=3.

$b= \frac{a+1}{a-1}$

It's easy to prove (by induction, for example) that the fraction on the right becomes smaller as a grows for nonnegative integers. Therefore, the answer will be found using the smallest a possible. a = 0 gives a negative b, a=1 gives a division by zero (it can be seen from the original equation that a=1 doesn't work), so the answer is a=2, b=3.

when i am put dy/ dx=1 the answer also true why i dont know anyway thats may be good luck only