# Number Theory Problem Watch

Announcements

This discussion is closed.

(Original post by

What do you mean, in white?

**J.F.N**)What do you mean, in white?

Galois.

0

Report

#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.

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.

0

Report

#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.

Contradiction, and n<=5.

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

(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.

Contradiction, and n<=5.

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

0

(Original post by

Consider squares modulo 3:

**dvs**)Consider squares modulo 3:

**0³=0, 1³=1, 2³=2, 3³=0 (mod 3)**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

Report

#25

(Original post by

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!

**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!

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).

0

(Original post by

You can't take an inequality and extend it to mod 3.

For example it's

The numbers mod n aren't ordered (in any meaningful algebraic way).

**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).

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

Report

#27

(Original post by

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.

**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.

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.

0

(Original post by

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.

**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.

Have you come up with a proof?

0

Report

#29

(Original post by

Have you come up with a proof?

**Galois**)Have you come up with a proof?

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.

0

Report

#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.

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.

0

Report

#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

Contradicition, and n<=5.

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

Contradiction, and n>=5.

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.

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

Contradicition, and n<=5.

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

Contradiction, and n>=5.

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.

0

Report

#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.

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

0

(Original post by

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.

**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.

Let a = 10 and b = 1 and you will see that the first inequality will hold.

0

(Original post by

Success!! See attached if you're interested, all typed up on latex.

**J.F.N**)Success!! See attached if you're interested, all typed up on latex.

Galois.

0

(Original post by

Success!! See attached if you're interested, all typed up on latex.

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

**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?

0

Report

#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

0

Report

#38

(Original post by

Just two comments:

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?

**Galois**)Just two comments:

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 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.

0

(Original post by

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.

**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.

Rep is yours

0

Report

#40

Ok, so let's start with another one as I promised

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.

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.

0

X

new posts

Back

to top

to top