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        riddles >> easy >> Commutative Square Roots?
        
(Message started by: THUDandBLUNDER on May 10^th, 2004, 2:50am)

Title: Commutative Square Roots?
Post by THUDandBLUNDER on May 10^th, 2004, 2:50am

Find all integers a,b,c such that [smiley=surd.gif][a + (b/c)] = a[smiley=surd.gif](b/c)

Are there any non-integer solutions?

Title: Re: Commutative Square Roots?
Post by towr on May 10^th, 2004, 4:26am

on 05/10/04 at 02:50:37, THUDandBLUNDER wrote:

Are there any non-integer solutions?

Should a,b,c all be non-integers, or just any one of them? (because in the latter case it's rather trivial, take a'=a, b'=b/c and c'=1, where a,b,c are an integer solution)

Title: Re: Commutative Square Roots?
Post by towr on May 10^th, 2004, 4:37am

::[hide]any c = b(a² - 1)/a [ne] 0 seems to solve it for any a and b[/hide]::

I suppose though that the 'commutative' in the title means there a '+' missing between a and [sqrt](b/c)
In which case the puzzle becomes somewhat harder.. ::)

Title: Re: Commutative Square Roots?
Post by Benoit_Mandelbrot on May 10^th, 2004, 5:59am

::[hide] We can solve sqrt(x+a)=a*sqrt(x) for x, and we have x=a/(a^2-1). This means that b=a, and c=a^2-1 when a is an integer. There are infinitely many except when a=±1. a, b, and c are integers. There should be an infinite amount of non-integer solutions when a is irrational. When a is rational, being n/m, then b=n*m and c=n^2-m^2, when c is not equal to zero. Only b and c would be integers. [/hide]::

Title: Re: Commutative Square Roots?
Post by Sameer on May 10^th, 2004, 12:49pm

How is this commutative?

Title: Re: Commutative Square Roots?
Post by towr on May 10^th, 2004, 1:13pm

on 05/10/04 at 12:49:30, Sameer wrote:

How is this commutative?

Magic :P

And in some sense a commutes into/out of the scope of the [sqrt]
Of course it doesn't make much mathematical sense (http://mathworld.wolfram.com/Commutative.html).

Title: Re: Commutative Square Roots?
Post by THUDandBLUNDER on May 15^th, 2004, 7:15pm

Quote:

How is this commutative?

It is the square root that is commuting! :P