In 2012, T. Miyazaki and A. Togb{e} gave all of the solutions of the Diophantine equations $(2am-1)^x+(2m)^y=(2am+1)^z$ and $b^x+2^y=(b+2)^z$ in positive integers $x,y,z,$ $a>1$ and $bge 5$ odd. In this paper, we propose a similar problem (which we call the shuffle variant of a Diophantine equation of Miyazaki and Togb{e}). Here we first prove that the Diophantine equation $(2am+1)^x+(2m)^y=(2am-1)^z$ has only the solutions $(a, m, x, y, z)=(2, 1, 2, 1, 3)$ and $(2,1,1,2,2)$ in positive integers $a>1,m,x,y,z$. Then using this result, we show that the Diophantine equation $b^x+2^y=(b-2)^z$ has only the solutions $(b,x, y, z)=(5, 2, 1, 3)$ and $(5,1,2,2)$ in positive integers $x,y,z$ and $b$ odd.
Let $alpha, beta in (0,1)$ such that at least one of them is irrational. We take a random walk on the real line such that the choice of $alpha$ and $beta$ has equal probability $1/2$. We prove that almost surely the $alphabeta$-orbit is uniformly distributed module one, and the exponential sums along its orbit has the square root cancellation. We also show that the exceptional set in the probability space, which does not have the property of uniform distribution modulo one, is large in the terms of topology and Hausdorff dimension.