An equation that takes Pythagoras to a new level | Hacker News Reader

An equation that takes Pythagoras to a new level

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140 pointssoheilpro22 comments

The linked paper https://fermatslibrary.com/s/proof-without-words-pythagorean... gives you the formula and I am much more a symbols guy than a geometry guy so I will prove it that way, first rearrange it:

(4T_n)^2=(4T_n+1)^2-(4T_n-1)^2+....(4T_n+n)^2-(4T_n-n)^2

Now, (4T_n+k)^2-(4T_n-k)^2=16T_nk (this, of course, shows up in the geometric proof as four rectangles where one side is 4T_n and the other is k)

so this is equivalent to

16T_nT_n = 16 T_n (1+...n)

which is equvialent to:

T_n=1+...n

But that is the very definition of T_n. Q.e.d.

Also if you want to take Pythagoras to a new level, Edsger W. Dijkstra who is much better known for his work in CS has a very interesting formulation https://www.cs.utexas.edu/users/EWD/transcriptions/EWD09xx/E... proving sgn(alpha+beta-gamma)=sgn(a^2+b^2-c^2) which includes and extends the Pythagoras theorem.

Wow, fermatslibrary has gone further downhill. It's almost unusable now with all the engagement stuff.

The caption "The equation 1⁰² + 1¹² + 1²² = 1³² + 1⁴², whose answer is that both sides equal 365, was immortalized in a different form in this 1895 painting: “Mental Arithmetic. In the Public School of S. Rachinsky.” (NIKOLAY BOGDANOV-BELSKY)" has funky exponents. Medium requires that I make an account to say that, so I am saying it here.

That formatting error is repeated in at least one other place too.

On the subject of that painting, can anyone tell me what the kids are wearing on their feet?

Muckluks. Or the 1890 Russian version

These are bast shoes [0], known in Russian as lápti, which were the typical footwear in villages until the 20th century. They are strapped to the leg so that they don't fall off.

[0]: https://en.wikipedia.org/wiki/Bast_shoe

I think the equation on the board is ( 10^2 + 11^2 + 12^2 +13^2 + 14^2 ) / 365 which solves for 2.

The students are expected solve this with "Mental Arithmetic". Try that in your head!

Thanks. I cannot see that on my screen. The powers of 1 caught my eye, though.

Happy Pi-Day!

Let a=12 then (a-2)^2 + (a-1)^2 + a^2 + (a+1)^2 + (a+2)^2 /365

and a binomial a+b squared yields a^2 + 2ab + b^2 so

a^2 + 4 + a^2+1 + a^2 + a^2+1 + a^2 + 4

add up all those a^2 's to get 5a^2 + 10

replace a by 12 and 720 + 10

       yields   730

Divide by 365

That's what the sharp kid in the foreground is doing at the moment of the painting.

Assuming one's memorized a table of squares already.

10^2 = 100

11^2 + 13^2 = 121 + 169 = 290

12^2 + 14^2 = 144 + 196 = 340

100 + 290 + 340 = 730

730 / 365 = 2

You can also "walk it up" if you haven't memorized them:

      10^2*5
    + (10+11)*4
    + (11+12)*3
    + (12+13)*2
    + (13+14)
    ------------

       500
     +  84
     +  69
     +  50
     +  27
     -----
       730

It's a bit of detail to juggle in your head but not too bad if you hang on to your current subtotal and remember where you are in the sequence to compute the next addend, and you know it's easier to add multi-digit numbers left-to-right than-right to-left.

You only carry from the ones place twice.

Ptolemy's theorem is another Pythagoras on steroids

https://m.youtube.com/watch?v=bJOuzqu3MUQ

I thought this was going to help me with Project Euler 261 [0], but it's not exactly the same, so no luck so far. Cool article anyway; linking geometry and algebra is always satisfactory.

[0] https://projecteuler.net/problem=261

How much CPU time and number theory do problems in that range generally require?

Brute force search of 10^10 (with some automatic monitoring of how wide a range of 'm' to scan at each candidate) is a lot but not a lot of a lot, if you code in a CPU efficient language and maybe use parallel computation and are willing to wait a while.

The difficulty and the required run time vary a lot starting from problem 150 or so. This one seems particularly difficult (it's rated at 85% difficulty, which is also the highest I've managed to solve). The problem here is that the search space is not 10^10, but higher, since if you iterate over k you don't know which values of m or n need to be tested. For m=1 it's clear that you can just use Pythagorean triples, but for bigger values I don't know if there is a reliable procedure to generate solutions.

Also 3^3 + 4^3 + 5^3 = 6^3

Beautiful.

Pythagoreans believed that the world was made of math -- and that the harmonies of mathematics revealed the harmonies of the universe. Pretty good for 6th century BC.

Medium has paywalled me. Any non-paywall links?

That is the reason I really don’t like Medium. So many click-bait articles and paywall

> Medium has paywalled me. Any non-paywall links?

The link works here in Firefox with NoScript, also in Dillo (no JS support).

Reply on news.ycombinator.com