HSC 2015 MX2 Marathon ADVANCED (archive) (2 Viewers)

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Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

 
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seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

By factorising the LHS we are forced to have x-y=1 and x+y=p.

So x=y+1 and p=2y+1.

So there are only solutions for odd primes, and for each odd prime p we have the unique solution pair ((p+1)/2,(p-1)/2).
 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

(Doesn't fit in the other thread)

 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

(Doesn't fit in the other thread)

x=0 tells you 0 is a root.

x=k for positive integral k tells you

k-1 is a root => k is a root.

The only poly with infinitely many roots is the zero poly.
 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

next question

Note that composing a reflection about 0 and a reflection about 1 (in that order) gives us the translation .

This means that for all real x.

In particular, .

Now, by a straightforward induction we have for (The constraint on x here ensures that in multiple applications of the rule we stay inside the interval in which this rule is valid.)

This implies

So unless , in which case .

Both situations are possible because the constant functions 1 and 0 satisfy the desired properties.
 

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Re: HSC 2015 4U Marathon - Advanced Level

 

SilentWaters

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Re: HSC 2015 4U Marathon - Advanced Level

Apply the AP-GP inequality for integers 1 to n:



Making the cancellation and moving the root to the other side, we obtain the result.
 

seanieg89

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Re: HSC 2015 4U Marathon - Advanced Level

Find all functions such that:

 

Sy123

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Re: HSC 2015 4U Marathon - Advanced Level

Find all functions such that:























----------

I don't think I made any mistakes, is that the solution you're looking for?
 
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