Tuesday 21 February 2012

A Pythagorean Curiosity

Today a colleague asked me to construct a nice question around Pythagoras' Theorem.

Toying with various ideas, I came up with the following figure:


Now it's easy enough to replace a, b, d and e with values, scrub c and one of the right angle symbols and then ask the students to show that the triangle must be right angled. But of course I wanted a nice set of integers for this question, so I wrote a little Python program to generate integer solutions to the set of equations:


Here's the first 30 solutions (given in the form {a, b, c, d, e}):

{9,12,15,20,25}, {16,12,20,15,25}, {18,24,30,40,50}, {25,60,65,156,169}, {27,36,45,60,75}, {32,24,40,30,50}, {36,48,60,80,100}, {45,60,75,100,125}, {48,36,60,45,75}, {49,168,175,600,625}, {50,120,130,312,338}, {54,72,90,120,150}, {63,84,105,140,175}, {64,48,80,60,100}, {64,120,136,255,289}, {72,96,120,160,200}, {75,180,195,468,507}, {80,60,100,75,125}, {81,108,135,180,225}, {90,120,150,200,250}, {96,72,120,90,150}, {99,132,165,220,275}, {100,240,260,624,676}, {108,144,180,240,300}, {112,84,140,105,175}, {117,156,195,260,325}, {125,300,325,780,845}, {126,168,210,280,350}, {128,96,160,120,200}, {128,240,272,510,578}

What stood out was that in most of these, e was a multiple of 5. When I isolated the instances where e was not a multiple of 5, I found that either a or b was!

My first version of my Python program only examined values of a and b up to 1000, so of course I ramped it up a little - to values up to 1000000 for both a and b.

And it still appears to be true. In every solution my program has generated, either e is a multiple of 5 or a or b is.

Looker closer at Pythagorean triplets in general, it is known that at least one number in a Pythagorean triplet has to be a multiple of 5. So if a and b are not multiples of 5, c must be. But if c is a multiple of 5, the Pythagorean triplet of {c, d, e} does not require that e also be a multiple of 5, and yet that appears to be the case for the figure at the top if it is to have integer sides where a and b are not multiples of 5.

Now if I could prove that this it always the case!

Update: when I showed this to another colleague, she pointed out the obvious fact that I had overlooked, namely that the two triangles in the diagram are similar.

Now the matter becomes simple.

Suppose that a is a multiple of 5. Given that the triangles are in the ratio a:c and a and c are the corresponding sides of the two triangles, we find that
and for e to be an integer would mean that c2 must be divisible by a and therefore divisible by 5, and therefore c must be. So both a and c will be multiples of 5. Now it's easy to show that if two of the numbers in a Pythagorean triplet are multiples of 5, the third must be a multiple of 5 as well, so we conclude that if a is a multiple of 5, so are b and c.

What if a is not a multiple of 5? As at least one number in a Pythagorean triplet must be a multiple of 5, if a is not, then either b or c is.

If b is the multiple of 5, c will not be. As stated before, the triangles are in the ratio a:c, so we find
and therefore d is also a multiple of 5. As c is not, we conclude that e is also not a multiple of 5.

The third possibility is that c is the multiple of 5. Again, noting that

but this time, where a and b are not divisible by 5, we conclude that e must be a multiple of 5 as well, and thus d must be as well.

So we find that the possibilities are as follows:
  • if a is a multiple of 5, b and c are multiples of 5 as well (and d and e may be as well)
  • if a is not a multiple of 5 but b is, then d is as well but c and e are not
  • if a and b are not multiples of 5, c must be, and d and e will be as well
Now I can sleep!