# Talk:Die Hard III

Joao Pedro Afonso Says: June 15, 2010 at 12:49 am | Reply

Saturday, someone said “This puzzle is probably unique in mentioning the numbers 3, 4 and 5 but having nothing to do with Pythagoras…”. I, of course, being a pleasure spoiler wrote to him Sunday, “actually, there is” one, even if smaller. It’s time to say what is the relation.

First of all, the numbers 3, 4 and 5 don’t have themselves a direct relationship with Pythagoras. This might sound weird because even last week, those numbers where instrumental to some of us to remember the right skills to reach the solution. But what Pythagoras is famous for, is for his theorem which relates the length of the sides A and B of any right triangles, with its hypotenuse C: A^2+B^2=C^2. This is valid for any right triangle whichever their sides might be, be integer values, fractional, rational or even irrational. There are always solutions C for any pair (A,B), but sometimes we might want all the solutions A, B and C, to be integer… in which case we say the equation is a Diophantine one. 3, 4, and 5 are famous because they are the simplest non-trivial solution to the equation… there are others, but these are the smaller ones. They are useful too: any carpenter knows that if they use rules of length 3 and 4 connected in one end, they can have a right angle if the distance between the other ends makes 5. This kind of knowledge probably predates Pythagoras, but he had the honor of having generalized it. What I’m saying is that, 3, 4, 5 are special because they are solution to a particular equation (used by Pythagoras). But, as it happens, these numbers lead to a solution of my kind, through the same properties. Just wait and see:

If we rewrite the Diophantine equation:

A^2+B^2=C^2 A^2=C^2-B^2 A*A = (C+B)(C-B)

What this mean is that, if the size of the jugs ‘C’ and ‘A’ and the quantity we want to have ‘B’, are integer solutions to Pythagoras Theorem, with the bigger jug ‘C’, 1 unit above the quantity ‘B’ we want, then a way to get B is to fill the jug A, A times, producing an excess B over C capacity. Let’s try with numbers of Richard’s puzzle:

3*3 = (5+4)(5-4)

Filling the 3 floz jug three times will produce 5 floz (the bigger jug) plus the wanted 4 floz. We can play with the formula to make more solutions to the Pythagoras theorem: A square of an odd number is easily broke in two consequent numbers, so, for instance, for 5 we have:

5*5=25=(12+13)*(13-12)

In other words 5^2+12^2=13^2

That means, if we had a 13 floz jug and a 5 jug, and wanted 12, filling the last five times and pouring it back in the greater jug, we would had the 12 floz we wanted. This is exactly the kind of solution I proposed which means, if the numbers are solutions to the Pythagoras theorem, plus an additional condition, then I know what kind of solution to propose. This is the relationship with the Pythagoras theorem… it is not a great relation, but I never said it was (actually I warned against such expectations). It is probably not as effective as @Oscar congruent technique, but was fun to write her.

The above is from a post on Richard Wiseman's blog.