The question:

Prove that for no integer n, n

^{6}+3n

^{5}-5n

^{4}-15n

^{3}+4n

^{2}+12n+3 is a perfect square

What should i read to solve problems like this?

hey guys! today I found a question paper that is almost four years old. Our teacher chose four of us for this Olympiad. there were a total of 10 questions with total 100 marks. I couldn't solve any of them. I still cant solve them. I want your insights for one of the question. if you want I can post others as well.

The question:

Prove that for no integer n, n^{6}+3n^{5}-5n^{4}-15n^{3}+4n^{2}+12n+3 is a perfect square

What should i read to solve problems like this?

The question:

Prove that for no integer n, n

What should i read to solve problems like this?

Last edited by S_r on Sat Jul 07, 2018 4:46 pm, edited 1 time in total.

- kenbrooker
**Posts:**113**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

And yes, I am interested in any other questions...

"*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."

Experience comes from Bad Judgment

Whatever the missing question is, its answer probably relies on the fact that that polynomial can be written as

(n-2)(n-1)n(n+1)(n+2)(n+3)+3.

Other one: suppose A1A2A3....An is an n sided regular polygon such that :

1/A1A2= 1/A1A3 + 1/A1A4. Determine number of sides of the polygon.

Use the fact that all squares are 0 or 1 modulo 4 (and non squares 2 or 3 mod 4).S_r wrote: ↑Sat Jul 07, 2018 3:38 amhey guys! today I found a question paper that is almost four years old. Our teacher chose four of us for this Olympiad. there were a total of 10 questions with total 100 marks. I couldn't solve any of them. I still cant solve them. I want your insights for one of the question. if you want I can post others as well.

The question:

Prove that for no integer n, n^{6}+3n^{5}-5n^{4}-15n^{3}+4n^{2}+12n+3 is a perfect square

What should i read to solve problems like this?

No. I cheated and typed it into Wolfram Alpha.

If I were in a competition, I would certainly have evaluated the polynomial (call it p(x) ) at a few small values of x, like 0, +-1, +-2, just to get a feel for what it is like, I would then have noticed the unusual fact that the result was 3 every time. That means that p(x)-3 has linear factors x, (x-1), (x+1), (x-2), (x+2) and one other.

- kenbrooker
**Posts:**113**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

Again,

without divulging my massive derivation,

I think the polygon has zero sides (of

any length : )...

(

"*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."

Experience comes from Bad Judgment

I don't know if what you think is right or wrong. unless you can prove it it ain't right. Until then I can't believe if has zero sides.

- kenbrooker
**Posts:**113**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

Far as I know, a "regular" polygon has sides of equal length,

call it length S, so:

1/A1A2 = 1/A1A3 + 1/A1A4 is the same as

1/S^2 = 1/S^2 + 1/S^2, the same as

1 = 1 + 1 so

No polygon

satisfies...

I could be missing something but

did you say there were

10 questions?

call it length S, so:

1/A1A2 = 1/A1A3 + 1/A1A4 is the same as

1/S^2 = 1/S^2 + 1/S^2, the same as

1 = 1 + 1 so

No polygon

satisfies...

I could be missing something but

did you say there were

10 questions?

"*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."

Experience comes from Bad Judgment

3-sided:

A1A2 = A1A3 = A2A3 = 1 = S

And if A4 didn't exist, A1A4 = infinity

-> 1/1 = 1/1 + 1/infinity ...

-> 1 = 1 + 0

The usage of capital letters is typical for Points, small letters are used for lengths.

A1A2 = A1A3 = A2A3 = 1 = S

And if A4 didn't exist, A1A4 = infinity

-> 1/1 = 1/1 + 1/infinity ...

-> 1 = 1 + 0

The usage of capital letters is typical for Points, small letters are used for lengths.

Akenbrooker wrote: ↑Mon Jul 09, 2018 8:21 amFar as I know, a "regular" polygon has sides of equal length,

call it length S, so:

1/A1A2 = 1/A1A3 + 1/A1A4 is the same as

1/S^2 = 1/S^2 + 1/S^2, the same as

1 = 1 + 1 so

No polygon

satisfies...

I could be missing something but

did you say there were

10 questions?

A

I take my thanks back

Last edited by S_r on Mon Jul 09, 2018 11:51 am, edited 1 time in total.

If n>3, there is no integer solution:

a = (n-2)/n * pi

1/1 = 1/(2-2cos(a)) + 1/(3-2cos(2a-pi))

-> n = 13.757

a = (n-2)/n * pi

1/1 = 1/(2-2cos(a)) + 1/(3-2cos(2a-pi))

-> n = 13.757

I can't still see it can you make it clear?

The answer is 7 but I don't know how

I can derive a cubic equation that cos(x) must satisfy where x is half the central angle, i.e. x=180/n.

It is straightforward to see that:

A1A2 = 2sin(x)

A1A3 = 2sin(2x)

A1A4 = 2sin(3x)

Substituting this in the equation, and using the angle sum formulae sin(a+b)=sin(a)cos(b)+sin(b)cos(a) and cos(2a)=2cos(a)^2-1 you eventually get 8z^3-4z^2-z+1=0, where z=cos(x).

You can easily verify with a calculator that n=7 works.

Actually proving it is harder. There is probably some clever way to do this with complex numbers.

It is straightforward to see that:

A1A2 = 2sin(x)

A1A3 = 2sin(2x)

A1A4 = 2sin(3x)

Substituting this in the equation, and using the angle sum formulae sin(a+b)=sin(a)cos(b)+sin(b)cos(a) and cos(2a)=2cos(a)^2-1 you eventually get 8z^3-4z^2-z+1=0, where z=cos(x).

You can easily verify with a calculator that n=7 works.

Actually proving it is harder. There is probably some clever way to do this with complex numbers.

So people here can't solve an Olympiad question

I probably could but I can't be bothered. Spending a couple of minutes here and there during breaks at work is rather different from actually solving it in Olympiad. Olympiad questions are often hard until some specific insight hits you, and it can take many failed attempts and hard work before you get that insight. For now, the only way I can see of proving it takes too much time. In a real Olympiad you have lots of questions, and you can choose which ones to tackle as no one gets them all.

- kenbrooker
**Posts:**113**Joined:**Mon Feb 19, 2018 3:05 am**Location:**Oregon, USA

Thanks!jaap wrote: ↑Mon Jul 09, 2018 9:02 amA_{1}, A_{2}, A_{3}, A_{4}are consecutive vertices of the polygon.

A_{1}A_{2}is presumably supposed to mean the length of the side from A_{1}to A_{2}, which you dubbed S. Note however that A_{1}A_{3}and A_{1}A_{4}represent the lengths of diagonals of the polygon, which will be larger than S.

"*Good Judgment comes from Experience;*

Experience comes from Bad Judgment..."

Experience comes from Bad Judgment