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#21
(Original post by J.F.N)
What do you mean, in white?
I mean like this.

Galois.
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14 years ago
#22
Here was my original method (I had a sligh aritmetic slip so ended up with n=4). Obviously it doesn't contain the answer, but maybe someone can use it?

a^2 + b^2 = nab - n
a^2 - 2ab + b^2 = (n-2)ab - n
(a-b)^2 = (n-2)ab - n
a^2 + 2ab + b^2 = (n+2)ab - n
(a+b)^2 = (n+2)ab - n
So both (n-2)ab - n and (n+2)ab - n are perfect squares that are 4ab apart.
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14 years ago
#23
Here's another solution I just worked up. I'm not really satisfied with it, so there maybe be some mistake somewhere, but who knows!

(a^2 + b^2)/(ab - 1) = n
Let n>5, then:
a^2+b^2 > 5ab - 5
a^2-2ab+b^2 > 3ab-3-2
(a-b)^2 > 3(ab-1) - 2
Consider squares modulo 3:
0³=0, 1³=1, 2³=2, 3³=0 (mod 3)
Also: 3(ab-1) - 2 = -2 = 1 (mod 3)
=> (a-b)^2 > 1 (mod 3)
But squares cannot be > 1 modulo 3.
Now let's assume n<5, then by the same method we get another contradiction. So both n>5 and n<5 are false. Then n must equal 5.
QED
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#24
(Original post by dvs)
Consider squares modulo 3:
0³=0, 1³=1, 2³=2, 3³=0 (mod 3)
You said consider squares, yet you considered cubes...

Also,

0^2 = 0, 1^2 = 1, 2^2 = 1, 3^2 = 0 (mod3)

So the first part of your proof holds, i.e. n <= 5

But when you revearse the situation, you get a square < 1 modulo 3, but that is possible, as you can see a square = 1 or 0 modulo 3.

So you've only shown that if an integer solution exists, then it can only be at most 5.

Galois.
0
14 years ago
#25
(Original post by dvs)
Here's another solution I just worked up. I'm not really satisfied with it, so there maybe be some mistake somewhere, but who knows!
You can't take an inequality and extend it to mod 3.

For example it's true that 3>1. But if you took this to mod 3 you'd get 0>1 and hence a contradiction - at least by your logic.

The numbers mod n aren't ordered (in any meaningful algebraic way).
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#26
(Original post by RichE)
You can't take an inequality and extend it to mod 3.

For example it's true that 3>1. But if you took this to mod 3 you'd get 0>1 and hence a contradiction - at least by your logic.

The numbers mod n aren't ordered (in any meaningful algebraic way).
Firstly you cannot use the example 3 > 1 because 3 is not a square.

Secondly, if you look at an expression (a^2) mod 3 it will never be greater than 1, because all squares mod 3 are either 1 or 0.

So saying a^2 mod 3 > 1 mod 3 arrives at a contradiction.

Thus, what he wrote was correct to a certain extent.

Galois.
0
14 years ago
#27
(Original post by Galois)
Firstly you cannot use the example 3 > 1 because 3 is not a square.

Secondly, if you look at an expression (a^2) mod 3 it will never be greater than 1, because all squares mod 3 are either 1 or 0.

So saying a^2 mod 3 > 1 mod 3 arrives at a contradiction.

Thus, what he wrote was correct to a certain extent.

Galois.
I can use 3>1 perfectly well - because all I am looking to do is to show that one cannot take an inequality of integers and turn it into a meaningful inequality of mod 3 (or mod n) numbers.

a^2 mod3 will never be greater than anything - simply because that sentence does not make sense! You might as well ask whether "i >0" is true in the complex numbers.
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#28
(Original post by RichE)
I can use 3>1 perfectly well - because all I am looking to do is to show that one cannot take an inequality of integers and turn it into a meaningful inequality of mod 3 (or mod n) numbers.

a^2 mod3 will never be greater than anything - simply because that sentence does not make sense! You might as well ask whether "i >0" is true in the complex numbers.
Lol I was just teasing you there mate. I realised before that my first reply to 'dvs' was wrong

Have you come up with a proof?
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14 years ago
#29
(Original post by Galois)
Have you come up with a proof?
I'm afraid I haven't looked at this.

It's the sort of thing that the IMO love to ask and the answers always take ages (at least for me) because they're so fiddly.
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14 years ago
#30
lol.. cubes. I have no idea what I was thinking. It's probably because I'm used to 'considering cubes modulo n'.

I also had a feeling we couldn't use modular arithmetic in inequalities -- that's why I said I wasn't satisfied with my solution.
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14 years ago
#31
Okay, here I go again:

Assume that (a^2 + b^2)/(ab - 1) > 5, then:
a^2 + b^2 > 5ab - 5
(a-b)^2 > 3ab-5
Let a=2, b=1:
(a-b)^2 = 1
3ab-5 = 1
Let (a^2 + b^2)/(ab - 1) < 5, then:
a^2+b^2 < 5ab-1
(a-b)^2 < 3ab-5
Let a=2, b=1:
(a-b)^2 = 1
3ab-5 = 1
So 5=<n<=5.
=> n=5
QED
<can it be this simple?! what am i missing?>

I've had a very busy day, and I'm quite tired. This is my last attempt at the question for tonight.
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14 years ago
#32
Wait, nevermind, that's not right.
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14 years ago
#33
Success!! See attached if you're interested, all typed up on latex.

Edit: I've also attached an addendum in the name of rigor.
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#34
(Original post by dvs)
Okay, here I go again:
...............
I've had a very busy day, and I'm quite tired. This is my last attempt at the question for tonight.
a = 2, and b = 1 is a trivial set of solutions for the expression to equal 5. So obviously the inequality will not hold, because the values you chose simply make it an equality.

Let a = 10 and b = 1 and you will see that the first inequality will hold.
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#35
(Original post by J.F.N)
Success!! See attached if you're interested, all typed up on latex.
J.F.N, I must rush to lectures so I will look at your work in a couple of hours.

Galois.
0
#36
(Original post by J.F.N)
Success!! See attached if you're interested, all typed up on latex.

Edit: I've also attached an addendum in the name of rigor.

The first line says that its impossible to have a = b, because this implies that

a^2 = -n/(2-n)

But you said this is absurd? Why must it be? There is no restriction saying that n < 2 (i.e. = 1). Did you mean to say that -n/(2-n) is not a perfect square?

Second comment:

You've reached a conclusion that no matter what integers a and b are, n < 6. Am I right? But take a = 6, b = 1, and you get n = 7.4 > 6.

Or did you mean that if n is an integer, it can be at most 5?
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14 years ago
#37
n must be greater than 2, which can prove easily, n -> infinite (if n is not integer) cuz if we let a = 1, so n = b^2+1/(b-1), which goes to infinite when b -> infinite
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14 years ago
#38
(Original post by Galois)

The first line says that its impossible to have a = b, because this implies that

a^2 = -n/(2-n)

But you said this is absurd? Why must it be? There is no restriction saying that n < 2 (i.e. = 1). Did you mean to say that -n/(2-n) is not a perfect square?

Second comment:

You've reached a conclusion that no matter what integers a and b are, n < 6. Am I right? But take a = 6, b = 1, and you get n = 7.4 > 6.

Or did you mean that if n is an integer, it can be at most 5?
On comment 1: That was rash. As you poitned out, -n/(2-n) is not a perfect square.

On comment 2: clearly i'm talking about (positive) integer values of n, or it wouldn't mean anything in my first step to say that an integer a must divide b^2 + n.
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#39
(Original post by J.F.N)
On comment 1: That was rash. As you poitned out, -n/(2-n) is not a perfect square.

On comment 2: clearly i'm talking about (positive) integer values of n, or it wouldn't mean anything in my first step to say that an integer a must divide b^2 + n.
Good stuff!

Rep is yours
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14 years ago
#40
If x, y , z, k is positive integers so that
x^2 + y^2 + z^2 = k(xyz +1).
Prove there are 2 positive integers a and b so that k = a^2 + b^2
That's my 1st one.
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