HSC 2014 MX2 Marathon ADVANCED (archive) (1 Viewer)

Stygian · Aug 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

aDimitri said:
It can be expressed in that form though.
$(\sqrt{17m^{2} + 4n^{2}} +\sqrt{17n^{2} + 4m^{2}} )(\sqrt{17m^{2} + 4n^{2}} - \sqrt{17n^{2} + 4m^{2}} )$

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 · Aug 13, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Stygian said:
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 · Aug 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

aDimitri said:
$\text{Given that m & n are both positive integers, show that }\\ 17m^{2} + 4n^{2} \text{ and } 17n^{2} + 4m^{2} \text{ cannot both be perfect squares}$

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 · Aug 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
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 · Aug 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
what

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.

dunjaaa · Aug 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Holy shit, if you don't SR this year, I would be really surprised...

glittergal96 · Aug 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

dunjaaa said:
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.

Fade1233 · Aug 15, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
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.

Completely agree with dunjaaa.

Sy123 · Aug 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
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 · Aug 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
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 · Aug 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
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

Stygian · Aug 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

aDimitri said:
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

where did you get it from btw?

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

aDimitri · Aug 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Stygian said:
where did you get it from btw?

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

got it from a mate.
and i honestly have no idea

Sy123 · Aug 16, 2014

Re: HSC 2014 4U Marathon - Advanced Level

$\\ $If$ \ a,b,c \ $are positive reals so that$ \ a^2 + b^2 + c^2 = 1 \ $, prove$ \\ \\ \frac{a^5+b^5}{ab(a+b)} + \frac{b^5+c^5}{bc(b+c)} + \frac{a^5 + c^5}{ac(a+c)} \geq 3(ab + ac +bc) - 2$

glittergal96 · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $If$ \ a,b,c \ $are positive reals, prove$ \\ \\ \frac{a^5+b^5}{ab(a+b)} + \frac{b^5+c^5}{bc(b+c)} + \frac{a^5 + c^5}{ac(a+c)} \geq 3(ab + ac +bc) - 2$

But if we put a=b=c=1, this reads 3 >= 7?

Sy123 · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
But if we put a=b=c=1, this reads 3 >= 7?

Oh wait I forgot to put in the condition

$a^2+b^2+c^2 = 1$

Editing it in now.

Stygian · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
But if we put a=b=c=1, this reads 3 >= 7?

hmmmm yeah i'm assuming there is some other condition to it?

edit: thanks sy

glittergal96 · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

Sy123 said:
$\\ $If$ \ a,b,c \ $are positive reals so that$ \ a^2 + b^2 + c^2 = 1 \ $, prove$ \\ \\ \frac{a^5+b^5}{ab(a+b)} + \frac{b^5+c^5}{bc(b+c)} + \frac{a^5 + c^5}{ac(a+c)} \geq 3(ab + ac +bc) - 2$

Ah right. Okay, in that case we have:

$LHS-RHS=2+\sum_{\textsf{cyc}}\frac{a^4+b^4+a^2b^2-ab(a^2+b^2)-3a^2b^2}{ab}$

$=2(1-(a^2+b^2+c^2))+\sum_{\textsf{cyc}}\frac{(a^2-b^2)^2}{ab}\geq 0.$

(Since the first term is zero, and the second is clearly non-negative.)

Sy123 · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

glittergal96 said:
Ah right. Okay, in that case we have:

$LHS-RHS=2+\sum_{\textsf{cyc}}\frac{a^4+b^4+a^2b^2-ab(a^2+b^2)-3a^2b^2}{ab}$

$=2(1-(a^2+b^2+c^2))+\sum_{\textsf{cyc}}\frac{(a^2-b^2)^2}{ab}\geq 0.$

(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:

$a^5 + b^5 \geq a^2b^2(a+b)$

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

And then things become quite clear after

Sy123 · Aug 17, 2014

Re: HSC 2014 4U Marathon - Advanced Level

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

$\\ 0 < u < 1 \\ $Let$ \ u_{n+1} = \frac{1}{u_n} + u \ $where$ \ u_1 = 1 + u \\ \\ $Show that$ \ u_n > 1 \ $for all integral$ \ n \geq 1$

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