Perfect square result

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apple2357

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I recently came across this pattern:

2^2+3^2+6^2 is a perfect square.

3^2+4^2+12^2 is a perfect square.

4^2+5^2+20^2 is a perfect square

I can prove this algebraically starting with this kind of idea:

n^2+(n+1)^2+n^2 (n+1)^2

Does anyone have a more elegant proof? Or perhaps a picture proof? Is this a well known result? I did a quick search but nothing much relating to it.
 
One tweak I would consider for an algebraic proof would be the following:

Rewrite it as \(\displaystyle \ n^2 \ + \ (n + 1)^2 \ + \ [n(n + 1)]^2.\)

The smaller factors are one apart, so I could make the following substitution:

\(\displaystyle Let \ \ n = (x - 0.5), \ and \ \ let \ \ (n + 1) = \ (x + 0.5) \ \ in \ \ the \ \ altered \ \ expression.\)

Then work with \(\displaystyle \ (x - 0.5)^2 \ + \ (x + 0.5)^2 \ + \ [(x - 0.5)(x + 0.5)]^2.\)
 
apple2357 said:
I recently came across this pattern:

2^2+3^2+6^2 is a perfect square.

3^2+4^2+12^2 is a perfect square.

4^2+5^2+20^2 is a perfect square

I can prove this algebraically starting with this kind of idea:

n^2+(n+1)^2+n^2 (n+1)^2

Does anyone have a more elegant proof? Or perhaps a picture proof? Is this a well known result? I did a quick search but nothing much relating to it.
Click to expand...
You haven't shown a proof yet! You need to show that n^2+(n+1)^2+n^2 (n+1)^2 is a perfect square.
Maybe you can factor that expression or multiply it out and show that what results is a perfect square.
 
apple2357 said:
I recently came across this pattern:

2^2+3^2+6^2 is a perfect square.

3^2+4^2+12^2 is a perfect square.

4^2+5^2+20^2 is a perfect square

I can prove this algebraically starting with this kind of idea:

n^2+(n+1)^2+n^2 (n+1)^2

Does anyone have a more elegant proof? Or perhaps a picture proof? Is this a well known result? I did a quick search but nothing much relating to it.
Click to expand...
I suppose your thinking is that, whatever proof you wrote, the way in which you decided what to square felt awkward. (It would help if you had shown your proof, so we could see what our proofs have to be more elegant than!)

How about this one: [math]n^2+(n+1)^2+n^2(n+1)^2\\=n^2+n^2+2n+1+n^2(n+1)^2\\=2n^2+2n+1+n^2(n+1)^2\\=1+2n(n+1)+n^2(n+1)^2\\=(1+n(n+1))^2\\=(n^2+n+1)^2[/math]
I've never seen this, that I know of, though perhaps the recreational math community is familiar with it.
 
Suppose that [imath]n\in\mathbb{N}^+~\&~F(n)=n^2+(n+1)^2+[n(n+1)]^2[/imath].
I would use induction to prove that for each positive integer [imath]n,~F(n)[/imath] is a perfect square.
What is [imath]F(1)=~?[/imath]
 
pka said:
Suppose that [imath]n\in\mathbb{N}^+~\&~F(n)=n^2+(n+1)^2+[n(n+1)]^2[/imath].
I would use induction to prove that for each positive integer [imath]n,~F(n)[/imath] is a perfect square.
What is [imath]F(1)=~?[/imath]
Click to expand...
I'm curious why you would use induction over factoring?
 
apple2357 said:
Here is my proof:

View attachment 34465
Click to expand...
That's exactly what I did first; I didn't recognize immediately what it would be that would have to be squared at the end, and I assume that is the step you would consider inelegant, as it requires either guessing or a little algebra to find the middle coefficient. But in fact, it is as elegant as a proof can be. (A proof doesn't have to show how one discovered it.)

So, knowing the answer, I was then able to see a way to get to it that would look logical.
 
Dr.Peterson said:
That's exactly what I did first; I didn't recognize immediately what it would be that would have to be squared at the end, and I assume that is the step you would consider inelegant, as it requires either guessing or a little algebra to find the middle coefficient. But in fact, it is as elegant as a proof can be. (A proof doesn't have to show how one discovered it.)

So, knowing the answer, I was then able to see a way to get to it that would look logical.
Click to expand...

Yes, that's exactly why i found it unsatisfactory. How would you factorise a quartic that doesn't have any real solutions? But yes, i looked for something squared.

I also used some of the numerical cases:

2^2+3^2+6^2 =7^2

3^2+4^2+12^2 = 13^2

4^2+5^2+20^2=21^2

so for example if n=2 ( first line) then 7 is generated by n(n+1) + 1 i.e. n^2+n+1
 
I have been trying to find a visual proof by thinking about areas of squares but have made no progress so far....
 
lookagain said:
One tweak I would consider for an algebraic proof would be the following:

Rewrite it as \(\displaystyle \ n^2 \ + \ (n + 1)^2 \ + \ [n(n + 1)]^2.\)

The smaller factors are one apart, so I could make the following substitution:

\(\displaystyle Let \ \ n = (x - 0.5), \ and \ \ let \ \ (n + 1) = \ (x + 0.5) \ \ in \ \ the \ \ altered \ \ expression.\)

Then work with \(\displaystyle \ (x - 0.5)^2 \ + \ (x + 0.5)^2 \ + \ [(x - 0.5)(x + 0.5)]^2.\)
Click to expand...
I cant see what this offers? Am i missing something?
 
I must say that your theorem is not presented in a way that suggests an intuitive proof. It is less constrained than what can be proved. The perfect square has a specific form. Moreover, a different form of presentation leads to a difference of squares and no need to deal with any quartic. Start here.

[math]\text {Prove } n^2 + (n + 1)^2 = \{n(n + 1) + 1\}^2 - \{n(n + 1)\}^2 [/math]
The proof is very straightforward.

[math]\text {Let } m = n(n + 1).\\ \therefore \ \{n(n + 1) + 1\}^2 - \{n(n+ 1)\}^2 = (m + 1)^2 - m^2 = \{(m + 1) + m\}\{(m + 1) - m\} = 2m + 1.\\ \therefore \ \{n(n + 1) + 1\}^2 - \{n(n + 1\}^2 = 2 * n(n + 1) + 1 = 2n^2 + 2n + 1.\\ \text {And } n^2 + (n + 1)^2 = n^2 + n^2 + 2n + 1 = 2n^2 + 2n + 1.\\ \therefore n^2 + (n + 1)^2 = \{n(n + 1) + 1\}^2 - \{n(n + 1)\}^2. \text { Q.E.D.}\\ \text {Corollary I: } n^2 + (n + 1)^2 + \{n(n + 1)\}^2 = \{n(n + 1) + 1\}^2.\\ \text {Corollary II: } n^2 + (n + 1)^2 + \{n(n + 1)\}^2 \text { is a perfect square.} [/math]
Congratulations on finding a nice theorem.
 
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Dr.Peterson said:
That's exactly what I did first; I didn't recognize immediately what it would be that would have to be squared at the end
Click to expand...
Since the coefficients in n^4 + 2n^3 + 3n^2 + 2n + 1 went 1 2 3 2 1 I thought about dice. This immediately gave me (n^2+n+1)^2. You get the same outcomes if you rolled two four-sided dice numbered 1, 2, 3 and 4.
 
Upon re-reading my post, I realized that I had missed a nuance. The theorem, as I revised it, and Corollary I apply if n is any real number. But the concept of a perfect square relates to non-negative integers. So I either implied an unnecessary restriction on the theorem and Corollary I or failed to specify a necessary restriction on Corollary II.

I apologize. As I have said many times, I am not a mathematician and so often lack the rigor required of real mathematicians.
 
Steven G said:
Since the coefficients in n^4 + 2n^3 + 3n^2 + 2n + 1 went 1 2 3 2 1 I thought about dice. This immediately gave me (n^2+n+1)^2. You get the same outcomes if you rolled two four-sided dice numbered 1, 2, 3 and 4.
Click to expand...
Please explain a little more here... I recognise the binomial coefficients but why' this immediately gave me ...'?
 
apple2357 said:
Please explain a little more here... I recognise the binomial coefficients but why' this immediately gave me ...'?
Click to expand...
What "12321" made me think of was the fact that 111*111=12321! That's closely related to his idea.

But since such a thought is not strictly part of a proof, and since you had evidently already found (somehow) that the answer was (x^2+x+1)^2, I looked for a way to arrive at that more algebraically.
 
Not sure if this is relevant to the thread, but:
  • Not all triplets [imath]a,b,c[/imath] for which [imath]a^2+b^2+c^2 = d^2[/imath] can be represented as [imath]n,n+1,n(n+1)[/imath]. E.g.: (1,4,8), (6,6,7).
  • Even when [imath]b=a+1[/imath] it is not necessary that [imath]c = ab[/imath]. E.g.: (8,9,12), (11,12,24).
 
apple2357 said:
I cant see what this offers? Am i missing something?
Click to expand...

\(\displaystyle Let \ \ n = (x - 0.5), \ and \ \ let \ \ (n + 1) = \ (x + 0.5) \ \ in \ \ the \ \ altered \ \ expression.\)

\(\displaystyle \ (x - 0.5)^2 \ + \ (x + 0.5)^2 \ + \ [(x - 0.5)(x + 0.5)]^2 \ = \)

\(\displaystyle x^2 - x + 0.25 \ + \ x^2 + x + 0.25 \ + \ (x^2 - 0.25)^2 \ = \)

\(\displaystyle 2x^2 + 0.5 \ + \ x^4 - 0.5x^2 + 0.0625 \ = \)

\(\displaystyle x^4 + 1.5x^2 + 0.5625\)

Look to see if the above trinomial is a perfect square.

Does \(\displaystyle \ 2(\sqrt{x^4})(\sqrt{0.5625}) \ = \ 1.5x^2 \ ?\)

Or, put another way, if \(\displaystyle \ \bigg( \dfrac{1.5x^2}{2}\bigg)^2 \ = \ 0.5625x^4, \ \) then you know it is.

It does, so the tweaked expression expression equals \(\displaystyle \ (x^2 + 0.75)^2.\)

Therefore, \(\displaystyle \ n^2 + (n+1)^2 + n^2 (n+1)^2 \ \) is a perfect square.
 
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