Note that z is a root. By Vieta jumping, the other root z′ satisfies z + z′ = 3xy and zz′ = x&hairsp;2 + y&hairsp;2. Thus since z is positive, z′ is also positive, we see that z′ = 3xy − z gives another solution. Now, WLOG, assume x > y, then take Since y > 0, 2 − 3y ≤ −1, so f(x) < 0. Since f(t) is an upward-facing parabola, that means min(z, z′&hairsp;) < x < max(z, z′&hairsp;). That means that we can construct three new solutions: (x, y, 3xy − z), (x, 3xz − y, z), and (3yz − x, y, z) and these are distinct. By our calculation above, exactly one of the three new solutions will have a smaller maximum element than (x, y, z) (and the other two larger). Thus we proceed in this way, reducing the maximum element each time (which is the essence of Vieta jumping). Since we are working with only positive integers, we must eventually stop, which means we reach a solution that has not all elements distinct. It's left for us to consider such a solution. WLOG assume x = y, then 2x2 + z2 = 3x2z. Thus x2 | z2 and x | z, so write z = ax. So we get So we see a|2 so a = 1 or 2. If a = 1 then we get (1, 1, 1) and if a = 2 then we get (1, 1, 2). And from (1, 1, 2) we get to (1, 1, 1) by taking (x, y, 3xy − z). Thus we see that starting from an arbitrary solution we eventually come to (1, 1, 1), and so these are all the solutions. Other properties Aside from the two smallest singular triples (1, 1, 1) and (1, 1, 2), every Markov triple consists of three distinct integers. The unicity conjecture states that for a given Markov number c, there is exactly one normalized solution having c as its largest element: proofs of this conjecture have been claimed but none seems to be correct. Odd Markov numbers are 1 more than multiples of 4, while even Markov numbers are 2 more than multiples of 32. In his 1982 paper, Don Zagier conjectured that the nth Markov number is asymptotically given by The error is plotted below. Moreover, he pointed out that , an approximation of the original