Perfect square result

[Steven G]

(1 x^2+ 1x^1+1x^0)^2 = 1x^4 + 2x^3 + 3x^2 + 2x^1 + 1x^0

Consider two 3-sided dice both numbered 0, 1 and 2 (I have no idea why I said 4-sided dice in my previous post).
Now if you look at the Cayley table for the sum of the two dice, you'll see that you can get a sum of 0 in 1 way, 1 in 2 ways, 2 in 3 ways, 3 in 2 ways and a 4 in 1 way.

Consider 1x^2+ 1x^1+1x^0:
1x^2 means there is 1 face with a 2 on it
1x^1 means there is 1 face with a 1 on it
1x^0 means there is 1 face with a 0 on it

Now consider (1 x^2+ 1x^1+1x^0)^2 = 1x^4 + 2x^3 + 3x^2 + 2x^1 + 1x^0:
1x^4 means there is 1 way to get a sum of 4
2x^3 means there are 2 ways to get a sum of 3
3x^2 means there are 3 ways to get a sum of 2
2x^1 means there are 2 ways to get a sum of 1
1x^0 means there is 1 way to get a sum of 0.

I gave a talk on dice a short while ago and this equation you came up with made me think of my talk and gave me the factoring quite quickly.

 

[JeffM]

I have been trying to find a visual proof by thinking about areas of squares but have made no progress so far....

Please take another look at my post 14 and 16. They lead to a fairly easy geometric proof.

Construct a square, ABCD, where the length of a side is [imath]n(n + 1) + 1 = n^2 + n + 1[/imath] units.

Obviously, that square can be decomposed into
[imath]n^4 + 2n^3 + 3n^2 + 2n + 1[/imath] unit squares.

On the line AB, mark as E the point [imath]n(n + 1) = n^2 + n[/imath] units from A.

On the line AD, mark as F the point [imath]n(n + 1) = n^2 + n[/imath] as units from A.

Construct perpendiculars at E and F. Let their intersection be G.

It is trivial to show that AEGF is a square that can be decomposed into
[imath]n^4 + 2n^3 + n^2[/imath] unit squares and that no point of that square lies outside of ABCD.

Therefore, the square ABCD can be decomposed into square AEGF plus a residual area.

The residual area necessarily has an area equal to

[math](n^4 + 2n^3 + 3n^2 + 2n + 1) - (n^4 + 2n^3 + n^2) = 2n^2 + 2n + 1 = n^2 + (n + 1)^2 \text { unit squares.}[/math]
By presenting the theorem as one equating a difference of squares to a related sum of squares, we get a quite intuitive geometric interpretation of the theorem..

 

[apple2357]

I can see what you are sort of saying...

i tried to interpret your thoughts in a numerical example first so I can get my head around it.

So we have 3^2+4^2+12^2 is a perfect square, in fact it is 13^2

So we could set about exploring the rewrite of the result:

3^2+4^2 = 13^2-12^2 as it makes use of the difference of two squares:





I can see the residual area ( orange bit above) comes to 25 which is indeed 3^2 +4^2 but i cant see how to construct two squares in the orange area.

 

[JeffM]

If you go back to my first post, # 12, I restated your theorem in terms of equating a difference of two squares and a sum of two related squares and then proved the restated theorem. That proof was totally algebraic and involved no quartics or other weirdness. Admittedly, no geometry was involved. (Frankly, I studied geometry in 1959-60 and remember almost nothing about it.) In # 12, I missed some nuances discussed in post # 14, but they did not involve anything fundamental in the algebra.

Then I got to thinking about your interest in a more geometric proof and utilizing the difference of squares approach.

The orange area I called the residual area and merely showed that it necessarily contained the exact number of unit squares needed to construct a square with n unit squares on a side and a non-contiguous square with n + 1 unit squares on a side. I did not make any effort to do the actual decomposition of the residual area and subsequent construction of the two squares. I merely showed that it is feasible.

 

[apple2357]

Thanks. I got you.