In our last post on this topic, we solved an equation of the form *Px* + *Qy* = *T*, where *x* and *y* are the variables and *P*, *Q* and *T* are integers. We used a standard algorithm to solve it and then found a much shorter method that used only a couple of steps of reasoning. The possibility exists that the simplicity of the reasoning solution is due to there being no solutions, so let’s try one that does have solutions: Find all positive integral solutions for (1) . Solving for *y* in terms of *x*, we get (2) . (3) Let . Solving for *x* in terms of *A*, we get (4) . (5) Let . Solving for *A* in terms of *B*, we get (6) . (7) Let . Solving for *B* in terms of *C*, we get (8) . Applying equations (8) and (7) to equation (6), we get (9) . Applying (8) and (5) to (4), we get (10) . Applying (10) and (9) to (2), we get (11) . *C* must be an integer for both *x* and *y* to be integers. If *C* < 0, then *x*< 0. If , then *y* < 0. Therefore *C* = 0 or *C* = 1, yielding *x* = 2, *y* = 15, or *x* = 7, *y* = 2.

Can we get this solution by just reasoning about the original equation, 13*x* + 5*y* = 101 ? If , then *y* < 0, so *x* < 8. If , then *x* < 0, so *y* < 21. There are fewer possibilities for *x*, namely *x* = 1 through *x* = 7. The only integral values of *x* in this interval that yield integers for *y* are 2 and 7, yielding *y* = 15 and *y* = 2, respectively.

With so few potential solutions, the deck still seems stacked for ease of finding solutions without doing the tedious work of deriving the parametric equations or the even more tedious trial-and-error work of testing a huge number of potential solutions. So next time we’ll compare these two methods when the number of potential solutions is more daunting.

*TO BE CONTINUED*