Prove the square of an odd multiple of 3 is the difference of two triangular numbers

[ZeHgS]

Hi guys, there is an old text book in Portuguese called Fundamentals of Arithmetic I'm studying from, but I can't find solutions online. I'm absolutely stuck on this problem:

"Prove the square of an odd multiple of 3 is the difference of two triangular numbers."

So I know an odd multiple of 3 is always in the form of

(3*(2c + 1)), so that squared is 36c² + 36c + 9

I also know that a triangular number is in the form of n(n+1)/2. Looking online I found the same problem but with an extra sentence saying "in particular, show that for any natural number n, so that [3(2n + 1)]²ˆ = t_9n+4 - t_3n+1."

I can do it by cheating and working backwards from there

t_9n+4 - t_3n+1 =ˆ [(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2
= (72n² + 72n + 18)/2 ˆ = 36c² + 36c + 9

But I can't for the life of me show that 36c² + 36c + 9 is [(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2. I'm stuck at this point (72n² + 72n + 18)/2. How could I possibly (even know to) decompose
(72n² + 72n + 18)/2
into
[(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2?

Hope I was clear, because I am confused myself. Thanks a lot!

 

[stapel]

...How could I possibly (even know to) decompose
(72n² + 72n + 18)/2
into
[(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2?

I think this is one of those instances when the "hint" in your other source reveals the usual solution method: You figure out the answer, and then work backwards. How you're supposed to answer this without that hint, I have no idea! :shock:

 

[ZeHgS]

I think this is one of those instances when the "hint" in your other source reveals the usual solution method: You figure out the answer, and then work backwards. How you're supposed to answer this without that hint, I have no idea! :shock:

Thanks a lot for your reply! It's been many years since I graduated high school and stopped studying math but I started to study it on my own recently for fun. I'm happy this isn't something completely obvious that I missed!

After working on this for maybe a few hours and using the example (which came from an unrelated source I looked up online, so it originally shouldn't have been used) I think I understood how to turn 36c² + 36c + 9 into many other possible examples of a "difference of 2 triangular numbers" and the thought process that would allow this without the example that was given. Due to my severe lack of mathematical skills, I doubt I would ever have been able to without it though hahaha. I went step by step, trying to make 36c² + 36c + 9 look like x(x + 1)/2 - y(y +1)/2.

First, I changed it into (72n² + 72n + 18)/2 to have the "/2"

Then, I realized I needed a "-" in there as well. Since a polynomial decomposes into this form (Xn + b)(Yn + a), and since Xn + b = Yn + a + 1, which is what we want in order to have a difference of two triangular numbers, I set Xn = Yn, which left me with b = a + 1.

So the polynomial needs to decompose into this (Xn + a)(Xn + (a+1)).

Since X is the same for both, I realized the 72n² would then need to be written as a difference of two square numbers and I tested it manually and found at least another possibility, "121n² - 49n²", as opposed to the example's "81n² - 9n²".

Then for the "C" and the "B" in An² + Bn + C, I don't know what they are called lol. The "C" needed to be such that

18 = C1 - C2
18 = a.b - c.d

And, since one needs to be exactly 1 larger than the other

b = a + 1
d = c + 1
18 = a(a + 1) - c(c + 1)

Also, a and c would need to be such that a.sqrt(A) + (a + 1).sqrt(A) = "A" for the "B" coefficient.

I then tested it out manually and found other possible values. The example uses c as 1, I just went up from there one by one. c = 3 is already a match, which makes a = 5

Then I got

(72n² + 72n + 18)/2 = (121n² + 121n + 30 - 49n² - 49n - 12)/2
(72n² + 72n + 18)/2 = (11n + 5)(11n + 6)/2 - (7n + 3)(7n + 4)/2
Let 11n + 5 = x and 7n + 3 = y, then
(72n² + 72n + 18)/2 = (x)(x + 1)/2 - (y)(y + 1)/2

Which proves it is a difference of two triangular numbers! Or have I made any mistakes? How common is this "testing it manually" in math? Is it a valid method of proof? It bothers me a bit, not sure why hahaha

 

[pythagorean314159]

examine this

Hi guys, there is an old text book in Portuguese called Fundamentals of Arithmetic I'm studying from, but I can't find solutions online. I'm absolutely stuck on this problem:

"Prove the square of an odd multiple of 3 is the difference of two triangular numbers."

So I know an odd multiple of 3 is always in the form of

(3*(2c + 1)), so that squared is 36c² + 36c + 9

I also know that a triangular number is in the form of n(n+1)/2. Looking online I found the same problem but with an extra sentence saying "in particular, show that for any natural number n, so that [3(2n + 1)]²ˆ = t_9n+4 - t_3n+1."

I can do it by cheating and working backwards from there

t_9n+4 - t_3n+1 =ˆ [(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2
= (72n² + 72n + 18)/2 ˆ = 36c² + 36c + 9

But I can't for the life of me show that 36c² + 36c + 9 is [(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2. I'm stuck at this point (72n² + 72n + 18)/2. How could I possibly (even know to) decompose
(72n² + 72n + 18)/2
into
[(9n + 4)(9n + 5)]/2 - [(3n + 1)(3n + 2)]/2?

Hope I was clear, because I am confused myself. Thanks a lot!

Rewrite 72n² + 72n + 18 as (81n² + 45n + 20) - (9n² + 9n + 2). Then, factor each trinomial in parentheses.

 

[ZeHgS]

Rewrite 72n² + 72n + 18 as (81n² + 45n + 20) - (9n² + 9n + 2). Then, factor each trinomial in parentheses.

Thanks for your reply! Like I said in my original post, I figured out that's what they do, but by using the other source's already decomposed example. However, I couldn't figure out the thought process that would allow someone who hadn't seen an example to solve this until yesterday, and I explain it in my second post with an example of another polynomial that can be created through such process.