HSC 2014 MX2 Marathon ADVANCED (archive) (2 Viewers)

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Stygian

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Re: HSC 2014 4U Marathon - Advanced Level

It can be expressed in that form though.


You would have to then show that a and b cant both be integers.
hmmm that's true and working from there would be no progress whatsoever from the original question

would using the property that two perfect squares multiplied together give a perfect square and then working towards showing the product isn't a perfect square lead to a solution? I'm actually horrible with latex so forgive me for not having a proper go at it (i'm also a bit stumped lol)
 

aDimitri

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Re: HSC 2014 4U Marathon - Advanced Level

hmmm that's true and working from there would be no progress whatsoever from the original question

would using the property that two perfect squares multiplied together give a perfect square and then working towards showing the product isn't a perfect square lead to a solution? I'm actually horrible with latex so forgive me for not having a proper go at it (i'm also a bit stumped lol)
Well assuming they are both perfect squares of integers a & b, their product would be a perfect square of ab. So you would be left attempting to show that ab is not at integer.
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

Note that 0 and 1 are the only quadratic residues mod 3. It follows that if 3 divides a sum of two squares x^2+y^2 it must divide x and y.

Consequently, the prime factorisation of any sum of two squares must contain 3 raised to an even power (a very easy induction).

But summing your two equations gives 21(m^2+n^2)=a^2+b^2. This is a contradiction, as it implies that one of (a^2+b^2) or (m^2+n^2) contains an odd number of factors of 3.

So such an (m,n,a,b) cannot exist.

Cool question! :)
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

Note that 0 and 1 are the only quadratic residues mod 3. It follows that if 3 divides a sum of two squares x^2+y^2 it must divide x and y.

Consequently, the prime factorisation of any sum of two squares must contain 3 raised to an even power (a very easy induction).

But summing your two equations gives 21(m^2+n^2)=a^2+b^2. This is a contradiction, as it implies that one of (a^2+b^2) or (m^2+n^2) contains an odd number of factors of 3.

So such an (m,n,a,b) cannot exist.

Cool question! :)
what
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

Is it wrong? Or which part is unclear?

A quadratic residue mod d just means a possible remainder if you divide a positive square by d. To find the quadratic residues mod d you just square the numbers from 0 through d-1 and take remainders upon division by d.

If two numbers that either have remainders 0 or 1 upon division by 3 add up to something divisible by three, then clearly those two remainders must be 0 (so both are divisible by 3 in other words).

The second line is true because if 3| x^2+y^2, then by the previous line 3|x and 3|y so 3^2|x^2 and 3^2| y^2 so 3^2|x^2+y^2. A simple induction shows that we can always move from an odd power of 3 dividing x^2+y^2 to the next power of 3 also dividing x^2+y^2.

The third line is true because the LHS is 3*7*(sum of two squares) and the RHS is (sum of two squares), which is impossible if the largest power of 3 that divides a sum of squares is always even (because we pick up an extra 3 on the LHS).

Sorry, I might have been too brief in my explanation.
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

Holy shit, if you don't SR this year, I would be really surprised...
Lol, thats very nice of you to say.

It would be cool if I got one, but I think a lot of it comes down to luck.
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

Is it wrong? Or which part is unclear?

A quadratic residue mod d just means a possible remainder if you divide a positive square by d. To find the quadratic residues mod d you just square the numbers from 0 through d-1 and take remainders upon division by d.

If two numbers that either have remainders 0 or 1 upon division by 3 add up to something divisible by three, then clearly those two remainders must be 0 (so both are divisible by 3 in other words).

The second line is true because if 3| x^2+y^2, then by the previous line 3|x and 3|y so 3^2|x^2 and 3^2| y^2 so 3^2|x^2+y^2. A simple induction shows that we can always move from an odd power of 3 dividing x^2+y^2 to the next power of 3 also dividing x^2+y^2.

The third line is true because the LHS is 3*7*(sum of two squares) and the RHS is (sum of two squares), which is impossible if the largest power of 3 that divides a sum of squares is always even (because we pick up an extra 3 on the LHS).

Sorry, I might have been too brief in my explanation.
I was expecting an answer that has more knowledge grounded in the 4U course

Its not really a fair question, but good solution nonetheless even though I understand very little of it
 

glittergal96

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Re: HSC 2014 4U Marathon - Advanced Level

I was expecting an answer that has more knowledge grounded in the 4U course

Its not really a fair question, but good solution nonetheless even though I understand very little of it
I tried to find a lower level solution for quite a while but had no luck. Does one exist aDimitri?

Thanks, it's really not that complicated. I just know what modular arithmetic is from reading some olympiad training resources online and that's all you need here.
 

aDimitri

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Re: HSC 2014 4U Marathon - Advanced Level

I tried to find a lower level solution for quite a while but had no luck. Does one exist aDimitri?

Thanks, it's really not that complicated. I just know what modular arithmetic is from reading some olympiad training resources online and that's all you need here.
I was hoping you guys could find one hence why i posted it here. i spent a couple of days on it couldn't get anything :p
 

Stygian

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Re: HSC 2014 4U Marathon - Advanced Level

I was hoping you guys could find one hence why i posted it here. i spent a couple of days on it couldn't get anything :p
where did you get it from btw?

are you sure there is a solution in the scope of the 4u course haha?
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

 
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Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

Ah right. Okay, in that case we have:





(Since the first term is zero, and the second is clearly non-negative.)
Yep that is correct, that is probably a "hidden" version of mine which is basically:



By considering the factorisation of a^5+b^5 = (a+b)....

And then things become quite clear after
 

Sy123

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Re: HSC 2014 4U Marathon - Advanced Level

A little hard for the lower level marathon but a little easier for this level:

 
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