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The Proof is Trivial!

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Classic science teachers :')

Well I was only every really interested in theoretical physics and 'the big questions' (sheldon-cooper style, standard) and I met some current student at my open day, one from maths and one from natsci, and they said that maths was 100% the best route into it and that even 'maths with physics' in my first year would hold me back!

That was the first step and since then I've found that the beauty and satisfaction in pure maths is immense and is completely different to everything else I've experienced. Besides, I'm not really fussed about money and I don't want to dedicate my life to coming up with things that makes peoples lives easier, so why not explore the cosmos/ mathematical universe! :smile:

Why did they say that?

I can understand why that is attractive. I'd enjoy the practical experiments as well as the theoretical side. Whilst the money isn't my ultimate desire, the fact that I'd enjoy both Chem Eng and Physics means that it does play a part in my decision.

Reply 621

james22

Original post by Jkn

Just remembered a really nice STEP question I did while ago and thought of this...

Problem 88*

P(N)

is defined as the largest product that can be made from positive integers that add up to N.

i.e.

P(N)=max(a_{1}a_{2}...a_{n}) : N= a_{1}+a_{2}+...+a_{n}

.

Prove the Goldbach Conjecture

Spoiler

Solution 88*

By the AM-GM inequality we have

(a_{1}a_{2}...a_{n})^{1/n} \leq \frac{a_{1}+a_{2}+...+a_{n}}{n}

a_{1}a_{2}...a_{n}=<(\frac{a_{1}+a_{2}+...+a_{n}}{n})^n

we have equality iff ai=aj for all i,j

hence

max(a_{1}a_{2}...a_{n})=(\frac{N}{n})^n

EDIT: just realised I need to maximise this as n is not fixed.

(edited 10 years ago)

Reply 622

Mladenov

Solution 89

Both

A

and

B

are prime numbers. Thus,

B=7 \times 6+1=43

, and

A=173

. Hence

N=347

Spoiler

Solution 88

Obviously, we have to partition into parts which are

< 5

. Notice that

2^{3}<3^{2}

, thus we can assume that we have at most two

2

's and the remaining number are

3

's (

2^{2}=2+2

).
If

N =3m

P(N)=3^{m}

.
If

N = 3m+1

P(N)=3^{m-1}.2^{2}

N =3m+2

P(N)=3^{m}2

(edited 10 years ago)

Reply 623

Jkn

Original post by james22

max(a_{1}a_{2}...a_{n})=(\frac{N}{n})^n

EDIT: just realised I need to maximise this as n is not fixed.

I just thought of an awesome way to adapt this :biggrin:

Spoiler

Original post by Mladenov

Solution 89

Both

A

and

B

are prime numbers. Thus,

B=7 \times 6+1=43

, and

A=173

. Hence

N=347

Spoiler

Correct

Am I right in saying you have obtained these numbers through systematic exhaustion? It would be great to find an elegant solution (I wonder if it exists...)

Solution 88

Obviously, we have to partition into parts which are

\le 5

. Notice that

2^{3}<3^{2}

, thus we can assume that we have at most two

2

's and the remaining number are

3

's (

2^{2}=2+2

).
If

N =3m

P(N)=3^{m}

.
If

N = 3m+1

P(N)=3^{m-1}.2^{2}

N =3m+2

P(N)=3^{m}2

Correcto, but why is the first part 'obvious'?

(edited 10 years ago)

Reply 624

james22

Original post by Jkn

Fuuuuuuck I just thought of an awesome way to adapt this :biggrin:

Spoiler

I think my entire solution is rubbish, I just re-read the question and noticed that the a_i have to be integers so my answer does not work.

Reply 625

Mladenov

Original post by Jkn

Correct

Am I right in saying you have obtained these numbers through systematic exhaustion? It would be great to find an elegant solution (I wonder if it exists...)

You are looking for prime numbers in arithmetic progressions. I can think of some analytic number theory estimations , but then the solution is far from elementary. This problems does not seem to be elegant, since we have to deal with primes, which are greater than

100

Original post by Jkn

Correcto, but why is the first part 'obvious'?

Simply because

2+3=5 < 6

.
But in general, the global maximum of the curve

x(a-x)

occurs at

\displaystyle x=\frac{a}{2}

. Hence for

a > 5

, we can have greater product by taking

b+ (a-b)

, where

\displaystyle b= \left[\frac{a}{2}\right]

(edited 10 years ago)

Reply 626

Jkn

Original post by james22

I think my entire solution is rubbish, I just re-read the question and noticed that the a_i have to be integers so my answer does not work.

But you can have

max(a_{1},a_{2}...a_{n}) \leq (\frac{N}{n})^n

. The solution I have in mind eliminates the need to intuition and formalises the result by linking it to

e

Reply 627

Jkn

Original post by Mladenov

100

Always a shame to use exhaustion though

Spoiler

But in general, the global maximum of the curve

x(a-x)

occurs at

\displaystyle x=\frac{a}{2}

. Hence for

a \ge 5

, we can have greater product by taking

b+ (a-b)

, where

\displaystyle b= [\frac{a}{2}]

Excellent. It's the use of a fact like that that I wanted :smile:

Spoiler

(edited 10 years ago)

Reply 628

Mladenov

Original post by Jkn

Always a shame to use exhaustion though

Spoiler

Absolutely. As you might have seen, I post problems with requite ideas, rather than brute force methods. :biggrin:

Original post by Jkn

Excellent. It's the use of a fact like that that I wanted :smile:

Spoiler

It is quite natural to look at

x(a-x)

, for it preserves the sum, and, in addition, you can vary the product. :tongue:

I am gonna leave problem 81 for somebody else. Yet, I have mind-blowing solution which uses abelian varieties.

(edited 10 years ago)

Reply 629

Jkn

Original post by Mladenov

Absolutely. As you might have seen, I post problems with requite ideas, rather than brute force methods. :biggrin:

It is quite natural to look at

x(a-x)

, for it preserves the sum, and, in addition, you can vary the product. :tongue:

Which is what makes a good problem. Mmm, hence the possible link to AM :smile:

Post some problems! A similar level of difficulty to 86 would be nice (or perhaps BMO2-level)
------------
Wonder if anyone recognises where I got this from

Problem 90*

Find the sum of all values of a such that the equation

(x^2-x+a+1)^2=4a(5x^2-x+1)

has 3 distinct real solutions.

(edited 10 years ago)

Reply 630

Mladenov

Original post by Jkn

Post some problems! A similar level of difficulty to 86 would be nice (or perhaps BMO2-level.

Problem 91*

Solve in integers

x^{5}-x^{3}=y^{3}z

, where

y,z

are prime numbers.

Problem 92*

Define

a_{1}=1

, and for

n > 1

a_{n}=n(a_{1}+..+a_{n-1})

. Then for every

n \equiv 0 \pmod 2

a_{n}

is divisible by

n!

.
Also, find all

n \equiv 1 \pmod 2

such that

a_{n} \equiv 0 \pmod {n!}

Reply 631

Jkn

Original post by Mladenov

Problem 91*

Solve in integers

x^{5}-x^{3}=y^{3}z

, where

y,z

are prime numbers.

Solution 91

Rearranging gives

x^3(x+1)(x-1)=y^3z

. The RHS has 4 prime factors and so the LHS must have 4 prime factors.

This implies one of the five terms in the product must equal 1. Noting that the expression must be positive (i.e. not equal to zero) we conclude that x=2.

Substituting in gives

y^3z=24=2^{3}.3

. As y and z are prime they must match up with these number exactly. Therefore the only solution is

(x,y,z)=(2,2,3)

Reply 632

james22

Original post by Jkn

Which is what makes a good problem. Mmm, hence the possible link to AM :smile:

(x^2-x+a+1)^2=4a(5x^2-x+1)

has 3 distinct real solutions.

Can a be complex? If not I don't think there are any a that make it have 3 distinct real solutions.

Reply 633

Doglamlico

Original post by Jkn

Problem 90*
Find the sum of all values of a such that the equation

(x^2-x+a+1)^2=4a(5x^2-x+1)

has 3 distinct real solutions.

Fairly sure this is too messy to be the right way of going about this question, but you never know:

Take over the RHS to give

(x^2-x+a+1)^2-4a(5x^2-x+1)=0

3 distinct real solutions means that

(x^2-x+a+1)^2-4a(5x^2-x+1) = (ax^2+bx+c)(dx^2+ex+f)

where

b^2 > 4ac

and

e^2 = 4df

.
Compare coefficients to produce a sufficient number of simultaneous equations, and then solve.

I've tried to do it on paper but got a bit lost after I found all of the equations/inequalities I could.

(edited 10 years ago)

Reply 634

Lord of the Flies

Solution 68

Note that

x\not\in\mathbb{Z}:\;[n-x]+[x]=n-1

hence letting

x_i\to 1-x_i:

Unparseable latex formula:

\begin{aligned}\displaystyle \int_0^1 \cdots \int_{0}^1 \left[\sum_{i=1}^n x_i\right] dx_1\cdots dx_n &=\int_0^1 \cdots \int_{0}^1 \left[n-\sum_{i] dx_1\cdots dx_n\\&=\int_0^1 \cdots \int_{0}^1 \frac{n-1}{2}\;dx_1\cdots dx_n\\&=\frac{n-1}{2}\end{aligned}

Solution 92

Observe that

\displaystyle 1+\big(1\cdot 1!+\cdots +n\cdot n!\big) =(n+1)!

by repeatedly factoring the LHS, then it is kid's stuff to show that

a_{n>1} = \frac{1}{2}n\cdot n!

and obviously the only odd

n

for which

n!|a_n

is when

a_n\neq \frac{1}{2}n\cdot n!

so for

n=1

Reply 635

Jkn

Original post by james22

Can a be complex? If not I don't think there are any a that make it have 3 distinct real solutions.

What makes you think that? :P

Original post by The Polymath

Fairly sure this is too messy to be the right way of going about this question, but you never know

There are many ways to do problems like these, this method does seem like it could get a little messy, though seems promising :smile:

------------------

Original post by Mladenov

Problem 92*

Define

a_{1}=1

, and for

n > 1

a_{n}=n(a_{1}+..+a_{n-1})

. Then for every

n \equiv 0 \pmod 2

a_{n}

is divisible by

n!

.
Also, find all

n \equiv 1 \pmod 2

such that

a_{n} \equiv 0 \pmod {n!}

Original post by Lord of the Flies

Solution 92

Observe that

\displaystyle 1+\big(1\cdot 1!+\cdots +n\cdot n!\big) =(n+1)!

by repeatedly factoring the LHS, then it is kid's stuff to show that

a_{n>1} = \frac{1}{2}n\cdot n!

and obviously the only odd

n

for which

n!|a_n

is when

a_n\neq \frac{1}{2}n\cdot n!

so for

n=1

Very nice. I took a somewhat different approach:

Solution 92 (2)

Clearly the statement is true for n=1 and 2. We then proceed to form an inductive argument.

Let

p=\frac{a_{n}}{n!}

, where p is a positive integer.

\frac{p(n+1)}{n}=\frac{(a_1+...+a_{n-1})(n+1)^2}{(n+1)!}=\frac{(a_1+...+a_{n-1})+n(a_1+...+a_{n-1})+n((a_1+...+a_{n-1})+n(a_1+...+a_{n-1}))}{(n+1)!}=\frac{a_{n+2}}{(n+2)!}

This implies

(n+2)!|a_{n+2}

if and only if

\frac{p(n+1)}{n}

is an integer. This happens in only two cases. Either n|(n+1) or n|p. The first case arises if and only if n=1 (dealing exclusively with the natural numbers, as specified). The second case implies that, if n|(n-1)! or (n-1)!|n, (i.e. n is non-prime), the inductive hypothesis is complete. On the other hand, if

gcd(n, (n-1)!)=1

(i.e. n is prime), then this would imply

n|(a_1+...+a_{n-1})

which cannot be true as by the definition of

a_{n}

and the assumption we made that n is co-prime to all proceeding positive integers.

Hence, when considering each case base, induction is complete only complete for n=2 and not n=1 (i.e. all even numbers, with the addition of 1).

An attempt to offer more insight into the inner working of the question (i.e. "why only even numbers plus one?") is to notice that the condition, with our particular relation(

a_1=1

), is only true for the base case two because it is the only such number that satisfies

(n-1)! < n

Reply 636

Zephyr1011

Original post by Jkn

Problem 88*

P(N)

is defined as the largest product that can be made from positive integers that add up to N.

i.e.

P(N)=max(a_{1}a_{2}...a_{n}) : N= a_{1}+a_{2}+...+a_{n}

.

Prove the Goldbach Conjecture

Spoiler

Solution 88

Assume that

a_k\geq5

. Then

a_k-2\geq3

, and so a different combination of integers, where

a_k

is replaced with

a_k-2

and

2

, will have a greater product as

2(a_k-2)=(a_k-2)+(a_k-2)>(a_k-2)+2=a_k

. Therefore

a_k<5 \;\forall k

a_k=4

, this is equivalent to a combination where

a_k

is replaced with 2 2s, as they have the same product and sum. So, WLOG, there must be an optimal combination with

a_k<4\;\forall k

. Clearly if

a_k=1

, a combination where another number was instead increased by 1 would have a greater product, therefore

a_k\neq1

. Therefore

a_k=2

3

. If there are 3 or more terms which are 2, then a combination where there are 2 3s, instead of 3 of those 2s, there will be a greater product as

2+2+2=3+3

and

2^3<3^2

. Therefore the optimal combination will have only 3s and 2s, and at most 2 2s.

Therefore, if

N=0 \pmod{3}

P(N)=3^{\frac{N}{3}}

, if

N=1 \pmod{3}

P(N)=4\cdot3^{\frac{N-4}{3}}

and if

N=2 \pmod{3}

P(N)=2\cdot3^{\frac{N-2}{3}}

Reply 637

Jkn

Original post by Mladenov

I am gonna leave problem 81 for somebody else. Yet, I have mind-blowing solution which uses abelian varieties.

(Only just saw you put this!)

Sounds awesome! Message me with it if you like (though it sounds as though it is far too sophisticated for me) :smile:

Btw, I haven't actually tried this problem yet, I just thought I would post it for the sake of the historical reference

Oooh just thought of an easier version that may interest people: evaluating the case whereby the cubes do not need to be distinct

Reply 638

Zephyr1011

Original post by The Polymath

Fairly sure this is too messy to be the right way of going about this question, but you never know:

Take over the RHS to give

(x^2-x+a+1)^2-4a(5x^2-x+1)=0

3 distinct real solutions means that

(x^2-x+a+1)^2-4a(5x^2-x+1) = (ax^2+bx+c)(dx^2+ex+f)

where

b^2 > 4ac

and

e^2 = 4df

I don't think this will work, as a quadratic must have either a single repeated root, or 2 real or complex roots, and it was explicitly stated that there must be 3 distinct real roots. I think that as you are supposed to take the sum of all values of a where 3 distinct real roots exist, there must only be a few special values of a where this is true, so you won't be able to factorise it. As to how you find these special values, I have no idea.

Reply 639

Doglamlico

Original post by Zephyr1011

My logic was that for there to be 3 distinct real roots of a quartic, the quartic must be equal to the product of two quadratics, one which has two distinct roots, and the other 1 repeated root, giving 3 in total.