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        riddles >> easy >> Do you recognize this identity?
        
(Message started by: Christine on Jan 24^th, 2013, 10:02pm)

Title: Do you recognize this identity?
Post by Christine on Jan 24^th, 2013, 10:02pm

Game: What general identity is exemplified by the following:

3(1^2 + 3^2 + 7^2) = 2^2 + 4^2 + 6^2 + 11^2
3(2^2 + 11^2 + 16^2) = 5^2 + 9^2 + 14^2 + 29^2

Title: Re: Do you recognize this identity?
Post by Grimbal on Jan 25^th, 2013, 6:30am

It would be
[hide]3(a²+b²+c²) = (b-a)² + (c-a)² + (c-b)² + (a+b+c)²[/hide]

Title: Re: Do you recognize this identity?
Post by Christine on Jan 25^th, 2013, 9:24am

on 01/25/13 at 06:30:12, Grimbal wrote:

It would be
[hide]3(a²+b²+c²) = (b-a)² + (c-a)² + (c-b)² + (a+b+c)²[/hide]

Nice! Can we play a game where one offers examples and ask to identify the identity?

Since Grimbal answered correctly, could you continue and ask us to identify an identity?

Title: Re: Do you recognize this identity?
Post by Grimbal on Jan 25^th, 2013, 5:10pm

well, I don't know...

sqrt(10-2*sqrt(5))/2 * sqrt(10+2*sqrt(5))/2 * sqrt(10+2*sqrt(5))/2 * sqrt(10-2*sqrt(5))/2 = 5

PS: a few more
sqrt(3) * sqrt(3) = 3
sqrt(2) * 2 * sqrt(2) = 4
1 * sqrt(3) * 2 * sqrt(3) * 1 = 6

Title: Re: Do you recognize this identity?
Post by Immanuel_Bonfils on Jan 27^th, 2013, 3:48pm

[hideb]sqrt(a-b)/2*sqrt(a+b)/2*sqrt(a+b)/2*sqrt(a-b)/2 = (a^2-b^2)/16[/hideb]

Title: Re: Do you recognize this identity?
Post by peoplepower on Jan 27^th, 2013, 6:37pm

I do not work with or encounter nontrivial identities too often. Nevertheless, there is one that stands out to me from perhaps four years ago as being the first of its kind that I learned: [I will add cases as required--maybe I am the only one who finds this nontrivial. :)]

1/6(7+1+1+1+1+1) = 2
1/4(12+4+6+2) = 6
1/8(20+2+2+4+10+4+4+2)=6
[hide]1/2(6+2)=4[/hide]

It should be noted that this is an identity with a single positive integer parameter, so [hide]the number-theoretic functions are the ones to look at.[/hide]

Title: Re: Do you recognize this identity?
Post by Grimbal on Jan 28^th, 2013, 1:34am

on 01/27/13 at 15:48:44, Immanuel_Bonfils wrote:

[hideb]sqrt(a-b)/2*sqrt(a+b)/2*sqrt(a+b)/2*sqrt(a-b)/2 = (a^2-b^2)/16[/hideb]

That is not what I had in mind. I added a few other examples.

Title: Re: Do you recognize this identity?
Post by peoplepower on Jan 28^th, 2013, 3:29am

on 01/25/13 at 17:10:27, Grimbal wrote:

This is the product of sines identity, here (http://mathworld.wolfram.com/Sine.html) it is equation (24).

Title: Re: Do you recognize this identity?
Post by Grimbal on Jan 28^th, 2013, 4:54am

You got it!

To solve this, you had to see the sines.... ;)

Title: Re: Do you recognize this identity?
Post by peoplepower on Jan 28^th, 2013, 2:42pm

Or at least one has to interpolate the circle of radius 2 on the complex plane.

I added two more cases to mine, one is hidden so treat that one as a hint.

Title: Re: Do you recognize this identity?
Post by Grimbal on May 18^th, 2013, 10:30am

The solution is here (http://en.wikipedia.org/wiki/Euler%27s_totient_function#Menon.27s_identity).

Title: Re: Do you recognize this identity?
Post by peoplepower on May 24^th, 2013, 2:05am

That is correct! I admit that I did not know it was called Menon's identity.

By the way, how did you find it?

Title: Re: Do you recognize this identity?
Post by Grimbal on May 24^th, 2013, 2:35am

Quite difficult. ;)

Title: Re: Do you recognize this identity?
Post by peoplepower on May 24^th, 2013, 2:51am

Ah, well the rules of the game allows you the opportunity to list a few examples of a very difficult identity.

Title: Re: Do you recognize this identity?
Post by Grimbal on May 24^th, 2013, 1:22pm

I recognized Euler's totient function (7->6, 12->4, 20->8). I first thought it had to do with the cycle lengths of the powers of k modulo n. But it didn't quite work out. So I googled the Euler totient function and found the matching identity. I didn't know about that identity before.

Title: Re: Do you recognize this identity?
Post by Annettagiles on Oct 15^th, 2014, 4:41am

the sum of base values in the brackets in the left side will be equal to the number present in the last of the right side.