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# BMO Banter watch

1. Hello,

I was just attempting a question, which you can find here. I think I've got a solution, but I was hoping that someone would be kind enough to check it
(Original post by Question 3, BMO-2, 1998)
Suppose x, y, z are positive integers satisfying the equation

and let h be the highest common factor of x, y, z.
Prove that hxyz is a perfect square.
Prove also that h(y-x) is a perfect square.
Attempt
Let , and so there is no positive integer greater than one that divides all three (Property A). Therefore

Therefore we find that .

Suppose that .

Let:
, such that
, such that
, such that

Let us show that .

Suppose . Note that . . Similarly .*

But due to Property A, must be one, and so . A similar argument goes for and .

We can also show quite easily that are also pairwise coprime: suppose two of them have a common factor greater than one; then this divides all three of , which is a contradiction.

Therefore, , and so . We can similarly show that .*

Therefore . Q.E.D. 1.

Going back to equation , we find that

And therefore . Q.E.D. 2.
I think that works, but I was most unsure at the two paragraphs I've ended with a small asterisk*.

Thanks in advance for any help
2. It all looks alright to me.
3. Thanks very much.
4. Hmm. I was trying question 4 of this paper, and I think I've sorted out the first part, and although I'm pretty sure the route I'm trying for the second part isn't the intended one (involving a little calculus), I'm not sure why it's not working. If anyone help out, I'd appreciate it.
(Original post by Question 4, BMO-2, 1998)
Find a solution of the simultaneous equations

in which all of x, y, z are positive and prove that it is the only such solution.
Show that a solution exists in which x, y, z are real and distinct.
First Part Attempt
For positive numbers , by AM-GM

But since these terms can't be negative, they must sum to zero, meaning they are all equal, and hence we find that is the only possibility for positive
Attempt at Second Part
If we consider the polynomial

, where , then are the roots of . For the roots to be distinct we need the turning points (two distinct ones) of this cubic to essentially be either side of the 'w-axis'.

So the turning points are at and we find that these are at (), and so we require the following two inequalities:

and

And after a lot of messy algebra (perhaps some of it is wrong), we end up with:

and

Then I attempt to combine the two inequalities and get

which seems to imply that we have distinct real roots to P(w) for any c > 6, c < -6. But this clearly isn't true, for example checking c = 7 and plotting the graph, there is only one solution (near 5).
Can anyone lend a hand? Or perhaps suggest an alternate method? I can't seem to spot a nice algebraic method for this part. Thanks again for any help!
5. Sorry, I misread earlier, deleted as soon as I realised. For the last part, guess a value for one of them (sensible guess!) and see whether you can find a solution for the others
6. (Original post by SimonM)
Sorry, I misread earlier, deleted as soon as I realised.
No problem!
For the last part, guess a value for one of them (sensible guess!) and see whether you can find a solution for the others
Cheers.
Spoiler:
Show
The solution works fine.
I'm wondering what you meant exactly by 'sensible'.
7. I'm also curious. To me the obvious "nice" guess was x=-1, but that ends up with y = -1, z = 11/2.
x = -2 gives 3 distinct roots, even though the other 2 roots aren't particularly nice.

Still, trial and error is a perfectly reasonable approach here.
8. (Original post by DFranklin)
To me the obvious "nice" guess was x=-1, but that ends up with y = -1, z = 11/2.
Indeed.
I'm also curious.
Hmm. Considering the route I originally went for, I 'cheated' and used a computer for a little bit: I decided to plot a graph of the cubic discriminant against the value of c (some horrible quartic I think), and it showed that we get three distinct roots if x+y+z is somewhere in between -7.5 and -8.5, or something like that....
9. One quick question on BMO-2 2009, whether the following method works
(Original post by Question 1, BMO-2, 2009)
Find all solutions in non-negative integers a, b to
Spoiler:
Show

But the plus-or-minus sign can't be a plus, otherwise , which is not possible (as we are taking positive square roots in the original equation), and so therefore

Therefore is a square multiple of 41 (as otherwise is not an integer), and so is , as the equation is symmetric, but we note that

Therefore we find that essentially.

So we end up with ordered pairs of solutions
10. I have checked too closely but the method looks fine
11. (Original post by SimonM)
I have checked too closely but the method looks fine
12. When I (first) did this your method of squaring repeatadly was my option but glancing at it now this is far quicker;

and by leting we are left with the far simpler problem c + d = 7
13. (Original post by DeanK22)
When I (first) did this your method of squaring repeatadly was my option but glancing at it now this is far quicker;

and by leting we are left with the far simpler problem c + d = 7
I noticed that as well, but I don't think that strictly proves that a and b are of the form ; it doesn't really show why they can't be a number not of that form at least, I don't think it does.
14. by squaring once you can see that either a and b have at least 1 common prime factor with an odd exponent (and possibly other prime factors that are not common with even exponents) or a and b are perfect squares with no common factors. Clearly it must be the former say a = c^2k and b=d^2k. The result follows
15. sorry for the mini hijack but can anyone give me hints on this.

find the minimum value of
all i've spotted is
Spoiler:
Show
that 5=2+3 12=2*2*3 so i did the binomial expansion of (2+3)^n but i couldn't see where to go. also n>=m

hints?
16. (Original post by Totally Tom)
sorry for the mini hijack but can anyone give me hints on this.
No worries
find the minimum value of
all i've spotted is
Spoiler:
Show
that 5=2+3 12=2*2*3 so i did the binomial expansion of (2+3)^n but i couldn't see where to go. also n>=m

hints?
Hmm. Well 12-5 = 7. And you should be able to show that for all m and n (I'm guessing we're talking positive integers here), so the question is really are there any pairs (m, n) such that ?

Further hint:
Spoiler:
Show
Consider the two cases and . In the first, can you show that m must be even (actually a multiple of 4)? Similarly, in the second, can you show that n must be even? Then after a little rearrangement, think difference of two squares.
17. (Original post by DeanK22)
by squaring once you can see that either a and b have at least 1 common prime factor with an odd exponent (and possibly other prime factors that are not common with even exponents) or a and b are perfect squares with no common factors. Clearly it must be the former say a = c^2k and b=d^2k. The result follows
Ah yes. This is true.
18. Hmm I was trying BMO-2, 1994, Q1, and I've met a block - I'm probably missing something obvious
(Original post by Question 1, BMO-2, 1994)
Find the first integer n > 1 such that the average of

is itself a perfect square.
Spoiler:
Show
Well the average of the first n squares is:

So we're essentially looking for solutions to the equation:

Note that n+1 and 2n+1 are coprime, and also that 2n+1 is odd, so we have either the following cases:

Let's dismiss the first case.

, but -1 is not a quadratic residue modulo 6, so this is not possible.

So

So either or , x and y are both odd.

Now let us dismiss the latter case: , but 2 is not a quadratic residue modulo 3, so this is not possible.

So essentially we are left to find the integer solutions of , and I can't seem to do this (other than x = 1, y = 1, which results in n = 1).

I suppose I could look down the route of a massive case bash involving , but is there a nice approach?
Any hints would be great
19. Just grinding through options is doable (and what I think most people who get it out under time pressure end up doing - it's what I did).

There's some discussion of this question here: http://www.thestudentroom.co.uk/show...=475982&page=2
20. that is a quadratic diophantine equation that you can solve. Have you come across quadratic diophantine equations before?

infact, here is a site with step-by-step method with the above problem;

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