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Topic: Do you recognize this identity? (Read 5200 times) |
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Christine
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Do you recognize this identity?
« on: Jan 24th, 2013, 10:02pm » |
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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
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| « Last Edit: Jan 25th, 2013, 9:19am by Christine » |
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Grimbal
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Re: Do you recognize this identity?
« Reply #1 on: Jan 25th, 2013, 6:30am » |
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It would be
3(a2+b2+c2) = (b-a)2 + (c-a)2 + (c-b)2 + (a+b+c)2
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| « Last Edit: Jan 25th, 2013, 6:30am by Grimbal » |
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Christine
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Re: Do you recognize this identity?
« Reply #2 on: Jan 25th, 2013, 9:24am » |
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on Jan 25th, 2013, 6:30am, Grimbal wrote:It would be
3(a2+b2+c2) = (b-a)2 + (c-a)2 + (c-b)2 + (a+b+c)2 |
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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?
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Grimbal
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Re: Do you recognize this identity?
« Reply #3 on: Jan 25th, 2013, 5:10pm » |
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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
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| « Last Edit: Jan 28th, 2013, 1:34am by Grimbal » |
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Immanuel_Bonfils
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Re: Do you recognize this identity?
« Reply #4 on: Jan 27th, 2013, 3:48pm » |
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| hidden: | | sqrt(a-b)/2*sqrt(a+b)/2*sqrt(a+b)/2*sqrt(a-b)/2 = (a^2-b^2)/16 |
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peoplepower
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Re: Do you recognize this identity?
« Reply #5 on: Jan 27th, 2013, 6:37pm » |
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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
1/2(6+2)=4
It should be noted that this is an identity with a single positive integer parameter, so the number-theoretic functions are the ones to look at.
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| « Last Edit: Jan 28th, 2013, 2:41pm by peoplepower » |
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Grimbal
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Re: Do you recognize this identity?
« Reply #6 on: Jan 28th, 2013, 1:34am » |
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on Jan 27th, 2013, 3:48pm, Immanuel_Bonfils wrote:| hidden: | | sqrt(a-b)/2*sqrt(a+b)/2*sqrt(a+b)/2*sqrt(a-b)/2 = (a^2-b^2)/16 |
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That is not what I had in mind. I added a few other examples.
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peoplepower
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Re: Do you recognize this identity?
« Reply #7 on: Jan 28th, 2013, 3:29am » |
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on Jan 25th, 2013, 5:10pm, Grimbal wrote: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
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This is the product of sines identity, here it is equation (24).
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| « Last Edit: Jan 28th, 2013, 3:30am by peoplepower » |
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Grimbal
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Re: Do you recognize this identity?
« Reply #8 on: Jan 28th, 2013, 4:54am » |
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You got it!
To solve this, you had to see the sines.... 
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| « Last Edit: Jan 28th, 2013, 4:57am by Grimbal » |
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peoplepower
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Re: Do you recognize this identity?
« Reply #9 on: Jan 28th, 2013, 2:42pm » |
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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.
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Grimbal
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Re: Do you recognize this identity?
« Reply #10 on: May 18th, 2013, 10:30am » |
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The solution is here.
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peoplepower
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Re: Do you recognize this identity?
« Reply #11 on: May 24th, 2013, 2:05am » |
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That is correct! I admit that I did not know it was called Menon's identity.
By the way, how did you find it?
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| « Last Edit: May 24th, 2013, 2:15am by peoplepower » |
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Grimbal
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Re: Do you recognize this identity?
« Reply #12 on: May 24th, 2013, 2:35am » |
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Quite difficult.
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peoplepower
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Re: Do you recognize this identity?
« Reply #13 on: May 24th, 2013, 2:51am » |
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Ah, well the rules of the game allows you the opportunity to list a few examples of a very difficult identity.
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Grimbal
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Re: Do you recognize this identity?
« Reply #14 on: May 24th, 2013, 1:22pm » |
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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.
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| « Last Edit: May 24th, 2013, 1:23pm by Grimbal » |
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Annettagiles
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Re: Do you recognize this identity?
« Reply #15 on: Oct 15th, 2014, 4:41am » |
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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.
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