I'm very confused with an a level maths q

• liz
In summary: More generally, all 3-tuples that are of the form (1, a, b), (1, b, c), (1, c, a), where ac+ b2= 1. And all 3-tuples of the form (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)), for natural m and n, are of the first form, as ac+ b2= (2 n2 + 2 n)2 + (2 n2 + 2 n +
liz
the question:
(im presuming n^2 means n squared)

prove the following result: "a traingle with sides that can be written in the form n^2+1, n^2-1, and 2n (where n>1) is right angled.
show, by means of a counter example, that the converse is false.

this q was taken from the back of the book "the curious incedent of the dog in the night-time" and there was a full proof but i don't understand some of it.

it start by exmplaining we need to prove which side is the longest by doing:

n^2+1 - 2n = (n-1)^2

if n>1 then (n-1)^2 >0

therefore n^2+1 > 2n

similarily (n^2+1) - (n^2-1) = 2

therefore n^2+1 > n^2-1

so n^2+1 > n^2-1.

does anyone understand this?

the rest is worked out using pythagoras but then the converse bit has completely lost me.

if anyone would like to explain, then i would be very grateful. thanks!

Last edited:
The converse would be:

If we have a right-angled triangle, then it's sides satisfy $$n^2 +1, n^2 -1, 2n$$. So you find a single particular example where this fails, and succeed in showing that the converse is not true.

Hope that helps, good luck.

Ro-me-o

thanks that's really helpful. do you understand the n^2+1-2n = (n-1)^2 ... stuff?

liz

liz said:
thanks that's really helpful. do you understand the n^2+1-2n = (n-1)^2 ... stuff?

liz

It's really just arithmetic- well, fancied up a little!

If a- b is a positive number, then a> b. Do you understand that?

To show that n2+ 1 is larger than 2n, look at the difference:
n2+ 1 - 2n.

That is equal to n2- 2n+ 1= (n- 1)2 and a square is never negative. (If n= 1, that can be 0 but if n> 1 it is positive and so n2+ 1 is larger than 2n.

Similarly, to show that n2+ 1 is larger than n2- 1 (which is almost obvious anyway) look at the difference (n2+1)- (n2- 1)= (n2- n2)+ (1- (-1))= 2 which is positive.

(The reason I said "almost obvious" is that it should be clear that n2+1> n2> n2-1.)

Once you know that, to use the Pythagorean theorem, look at
(n2+1)2= n4+ 2n2+ 1

(n2-1)2= n4- 2n2+ 1
(2n)2= 4n2

so (n2-1)2+ (2n)2= n4+ 2n2+ 1. Yep, c2= a2+ b2 so this is a right triangle.

The converse would be "If a, b, c are the sides of a right triangle (with a,b,c integer valued?), c being the hypotenuse, then c= n2+ 1, a= n2-1, b= 2n for some integer n. My first thought was the simple 3,4,5 triangle, but that turns out to work: 3= 22-1, 4= 2(2), 5= 22+ 1. Okay, what about 6, 8, 10?

c = 13 is a counterexample.

thank you it looked more difficult than it is. but in my defence i showed it to my boyfriend who is just finishing his maths A level and he couldn't understand it either. a counter example that works is 65, 60 and 25.

liz said:
thank you it looked more difficult than it is. but in my defence i showed it to my boyfriend who is just finishing his maths A level and he couldn't understand it either. a counter example that works is 65, 60 and 25.
I wish I knew how to say this, but I think the set of all 3-tuples (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)), for natural m and n, will give you the lengths of the sides of all right triangles with sides of natural length. And I think n = 1 is the only value for which (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1) = (x2 - 1, 2x, x2 + 1), for some natural x > 2. So every element of the set of all 3-tuples (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)), for natural m and n, with n > 1, should be a counterexample. Just some casual observations - maybe someone else could try to prove it.

i don't know i don't really understand how you got that probably cos i don't know what "3- tuples" or "natural m and n" means. i didnt actually think of the counter examle myself. the question with the answer including the counter example was taken from a book. i worked out that the 65, 60 and 25 triangle is a right angle using pythagoras then used 65= n^2+1 to get n=8. then the other sides would be 8^2 -1 = 63 and 2n = 16. which isn't 60 and 25 therefore proving that not all right angled have sides n^2+1, n^2-1 and 2n.
or so i thought. I am probably wrong, I am a bit out of my depth with this question.

liz said:
i don't know i don't really understand how you got that probably cos i don't know what "3- tuples" or "natural m and n" means.
Don't worry, it's simple. You already know the natural numbers; They're the members of the infinite set {1, 2, 3, 4, ...}. A tuple is just a finite sequence of objects, like (1, 2, 1). Order matters for sequences, so (1, 2, 1) is not the same as (2, 1, 1). There are x many objects in an x-tuple. So a 3-tuple is a sequence of 3 objects, a 10-tuple is a sequence of 10 objects, etc.
i didnt actually think of the counter examle myself. the question with the answer including the counter example was taken from a book. i worked out that the 65, 60 and 25 triangle is a right angle using pythagoras then used 65= n^2+1 to get n=8. then the other sides would be 8^2 -1 = 63 and 2n = 16. which isn't 60 and 25 therefore proving that not all right angled have sides n^2+1, n^2-1 and 2n.
or so i thought. I am probably wrong, I am a bit out of my depth with this question.
You're correct- your counterexample is actually my counterexample in a way- they form similar triangles (meaning, among other things, the lengths of their sides are proportional). My example was (5, 12, 13)- multiply each term by 5 to get your example.
I'm trying to prove the first part of my suggestion in another thread. I'll post my result, in case you're interested.

yes i am interested, thank you.

I was wrong. Though (m(2n + 1), m(2n2 + 2n), m(2n2 + 2n + 1)) always gives you the sides of a right triangle, it doesn't give you the sides of all right triangles. You can read all about the solution I was looking for here and here. (BTW, the 3-tuples I've been talking about are called Pythagorean Triples, and I discovered they have a long and interesting history.)

1. What is the question that you are confused about in A Level Maths?

The specific question that you are struggling with in A Level Maths can vary, but it is important to identify the topic or concept that you are having trouble with. This will help in finding the right resources or seeking assistance from a teacher or tutor.

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A Level Maths can be challenging for many students because it involves complex concepts and requires a strong understanding of fundamental mathematical principles. It also requires a lot of practice and critical thinking skills.

3. How can I improve my understanding of A Level Maths?

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Yes, it is normal to feel confused at times in A Level Maths. It is a challenging subject and it is common for students to struggle with certain topics or questions. The key is to not get discouraged and keep working towards understanding the material.

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