Diophantine footy

Last weekend, while watching my son’s footy team (GO BEARS!), my dad noticed that the Bears’ score, 7-7-49, had an unusual property, the score (calculated as 6 times the number of goals g plus the number of behinds b) was also the product of the goals and behinds. He posed the question: ‘Are there any other scores that have this property?’ Off the top of my head, it struck me that the score of 0-0-0, also satisfies the equation.

The next day, my dad emails me with two other solutions 4-8-32 and 3-9-27, and posed the question: ‘Are these all the solutions, and how would you prove that you have them all?’ So here it is:

So we are looking for cases when 6g + b = gb implies that g = b/(b - 6) (where g is the number of goals and b is the number of behinds). Note that we need g, b >= 0 and g, b whole numbers (hence Diophantine).

Also note that we need to have 2(b-6) <= b implies b \le 12 because otherwise g would not be a whole number, this limits the number of possible solutions. Also also note that the right hand side has b-6 in the denominator, so for g to be positive (and finite), we need b>6.

We can therefore proceed by brute-force and check all allowed values of \(b\):

b =  7, g = b/(b-6) =  7/1 = 7
b =  8, g = b/(b-6) =  8/2 = 4 
b =  9, g = b/(b-6) =  9/3 = 3 
b = 10, g = b/(b-6) = 10/4 
b = 11, g = b/(b-6) = 11/5 
b = 12, g = b/(b-6) = 12/6 = 2

Which gives us the solutions noted above, plus one new one 2-12-24 (the cases b=10,11 don’t provide us with whole goals). And that’s it. QED :)

Here is a plot of goals versus behinds, where you can see the integer solutions:

(Bonus: Check out the extra solutions if you allow anti-goals!)

(I think) there are actually a surprising number of solutions here. How would this work out if a goal was worth a different number of points (5, 7, 8, …)?