Thread: Pythagorean Theorem Question Problem

So we have been working on the Pythagorean Theorem and in this weekend's homework the following question was on the sheet.

"The lengths of the three sides of a right triangle are consecutive integers. Find them."

There was no other info. What does the question mean? What are "consecutive integers"?

I don't want anyone to do my homework, I just want help figuring out what this question means.

consecutive integers is any set of three numbers where the difference between each sucessive element is equal to one.

ie (1,2,3) or (4,5,6) or (21000 ,21001 ,21002) etc.

hence they want you to work out the lengths of the three sides of your triangle such that the lengths of the sides is one unit longer, or possibly shorter, than the previous length you used.

and of course the Pythagorean Theorem must still apply.

And a big hint is that there is only one Pythagorean triple made up of consecutive integers. mmm, so what is the standard pythagorean triple everyone knows?

Obviously not everyone knows...

I'll help narrow it down, the highest of the three numbers is less than 25.

It's always ued as an example so if you were to use a search engine.....

And my pythagorean triple of the day is (21, 220, 221)

interesting... let me think first

and todays pythagorean triple is (231, 520, 569)

There's also 33 44 55 - 333, 444, 555 - 3333,4444,5555 - etc etc.. :wink:

Originally Posted by river_rat

And a big hint is that there is only one Pythagorean triple made up of consecutive integers. mmm, so what is the standard pythagorean triple everyone knows?

Ya, well, I think you are being a bit unfair to the OP. Suppose, for example, you dont know the obvious Pythogorean triple referred to (most house builders do, by the way), how would one deduce it?

Other than trial and error, I cannot see how it could be done. Is there a way? I don't see it yet.

You don't need trial and error. Three consecutive integers look like n, n+1, n+2 for some integer n. Use the condition these are a Pythagorean triple.

Jeez, that was dumb of me, just get it in quadratic form, thus:

n² + (n + 1) ² = (n + 2)²

2n² + 2n + 1 = n² + 4n + 4

n² - 2n - 3 = 0

and solve for n using the quadratic equation formula we learned in school

The same trick shows that there is only one triple consisting of consecutive integers

Well, technically, there are 2 roots to the quadratic, but of course only one of them yields a right triangle, which is what the OP was concerned with.

Does anybody think there might be another universe out there where you can make a rightangled triangle with sides 5,5,5 ? :-D

What on Earth are you talking about? The roots of n² - 2n - 3 = 0 are 3 and -1. Where does 5 come into it? Have I missed your point?

Yes you have missed it, pythagorus works in our universe but is there another where a perfect r.a.t. might be 5,5,5 ?

it was err... humor..

Ya, not feeling humourous today, sorry old bean.

In fact there is a universe where 5, 5, 5 makes a right triangle, it's called a non-Euclidean manifold. Here, the internal angles of a triangle sum to more than 180 degrees, so I'm not sure what a right triangle means here. But. It just so happens we inhabit such a manifold, it's called space-time.