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You missed out one (or more) solution(s), because you (wrongly) assumed that x, y have to be integers: a = 18, b = 1 ==> a^b - b^a = 18^1 - 1^18 = 18 - 1 = 17, x = a^(b/2) = 18^(1/2) = sqrt(18) = 3 sqrt(2), y = b^(a/2) = 1^(18/2) = 1.
In the text there is not written a,b integer. If a,b integer, a,b are not both even or odd because 17 is odd then a is odd and b is even or viceversa. Equation is a^b=17+b^a then a^b≥17 and a≠b. We have to try even powers of an odd number. For a=3, powers are 9,,81, 729.............. But 9
A few of this break. Firstly a and b didn't start as integers. And even if they were, x = a^(b/2) may not be integers, and similarly with y. Therefore (x+y)(x-y) = 17 does not necessarily means either of them equating to 1 or 17.
Of course, it could have been shorter, but it is a presentation of the variety of tools that can be used to reach a solution. Naravno da je moglo i kraće ali u pitanju je prikaz raznovrsnosti alata koji se mogu koristiti da bi se došlo do rešenja.
it's missing the clauses, such as a and b are positive integers, and by "simplifying", you're breaking the function... a^b-b^a can be seen as a 3D function, and you're looking for all the points where your "surface" intersects with the flat plane at 17... and there can be a lot of "points", there.
In the video, the solution a = 18, b = 1 is lost when for the equation (x + y) * (x - y) = 17 the factors (x + y) and (x - y) are assumed to be integers. Using a = 18, b = 1, the equation is also fulfilled with the non-integer factors x = 18^(1/2) = sqrt(18), y = 1^(18/2) = 1 thus giving (x + y) * (x - y) = (sqrt(18) + 1) * (sqrt(18) - 1) = sqrt(18)^2 - 1^2 = 18 - 1 = 17.
If this is the way you're going to solve it - assume or try ...17x1, 3 and 4 ... then the best solution was to try 3 and 4 right from the beginning and QED
