Let $a,b$ be positive integers. Prove that if for all integers $s,t$ we have $sa+tb \neq \gcd(a,b)$ or $s+t=1$ then we have $a=b$ Question: Let $a,b$ be positive integers. Prove that if for all integers $s,t$ we have $sa+tb \neq \gcd(a,b)$ or $s+t=1$ then we have $a=b$.
Attempt:
Well I think for any integers $a,b$ by Bezout's theorem there must exist integers $s,t$ with $sa+tb=\gcd(a,b)$. So then $s+t=1$.
Then $s = 1-t$ and so
$(1-t)a + tb = \gcd(a,b) \implies a - t(a-b) = \gcd(a,b) $
then $t(a-b) <a$ since the gcd is positive.
From here I'm not really sure how to proceed.
Edit: Now that I actually understand this problem, the problem is saying is all of the Bezout coefficients $s,t$ have difference $1$ then $a=b$. To show this you can use the general form of the Bezout coefficients to choose two different representations and reason about $a-b$.
 A: Since you asked about the "thought process" in Igor's answer, I will show how we can view it simply as a special case of the well-known fact that lines are uniquely determined by two distinct points on the line. In particular this means that any two lines through two distinct points are identical, so have equal slopes.
By hypothesis all $(\ge 2)$ integer points $\,(x,y)\,$ on $\, a y = -bx + d\,$ also lie on the line $\,y  = -x + 1\,$ so these lines are identical, so they have equal  slope, i.e. $\,-b/a = -1,\,$ so $\,b = a,\,$ QED.
The proof in Igor's answer amounts to equating the slopes $\,m,m'\,$ of each line  computed from their two common points $(x,y)$ and $(x',y') = (x+a,y-b),\,$ yielding
$$ m = \dfrac{x'-x}{y'-y} = m'$$
Remark $ $ We don't need to known the values of the two common  integer points $\,(x,y),\,(x',y')\,$ on the lines - only that at least two such points exist. But general linear theory tells us infinitely many integer solutions exist for $\,a y +b x = d,\,$ viz. the general solution is the sum of any particular solution plus the general sum of the associated homogeneous equation $\,ay + bx=0.\,$ A particular solution exists by Bezout's identity for the gcd, and the homogeneous equation has an obvious solution $(x,y) = (a,-b),\,$ so adding it to any particular solution yields a second solution.
If you know linear algebra you may find it instructive to rephrase it in that language, e.g. using the fact that the determinant $\,m'-m\,$ of the system $\, y = m x + b,\ y = m' x + b'\,$ must be zero if the solution is not unique.
A: Note that if $t a + s b = (a, b),$ then $t_1 = t-b$ and $s_1=s+a$ also work. But $1 =  t_1 + s_1 = t + s + (a-b) = 1 + (a-b),$ so $a-b=0.$
