The Proof is Trivial!

Original post by Jkn

You've used the same method as mladenov, but I do find that the answer, when written in modular arithmetic form, is more aesthetically pleasing.

Do share some problems! :smile:

Oh, sorry, I hadn't seen their solution.

I will now see if I can find any interesting problems.

Edit: Ah, found one. The only problem I managed to get in the BMO2

Problem 93*

Prove there are an infinite number of positive integers m and n, such that

m|n^2+1

and

n|m^2+1

(edited 10 years ago)

Reply 642

Problem 94**

One really nice functional equation.

Find all strictly increasing functions

f: \mathbb{R} \to \mathbb{R}

such that

f(x+f(y))=f(x+y)+1

, for all real numbers

x

and

y

.

Problem 95** warm-up

Let

a,b,c

be positive real numbers. Then,

\displaystyle \sum_{cyc} \frac{a^{2}b^{2}(b-c)}{a+b} \ge 0

.

Problem 93 has been already posted.

(edited 10 years ago)

Reply 643

Original post by Zephyr1011

Oh, sorry, I hadn't seen their solution.

I will now see if I can find any interesting problems.

What I've been doing in some cases is trying to think of some myself; often generalisations of other problems I have tried/found :smile:

Such as...

Problem 96*

Let x, y and a be positive integers such that

x^3+ax^2=y^2

. Given that

1 \leq x \leq b

where b is a positive integer, find, in terms of a and b, the number of possible pairs

(x,y)

that satisfy the equation.

(edited 10 years ago)

Reply 644

Original post by Zephyr1011

Prove that

\forall n\in\mathbb{N}

, there exists n consecutive integers such that none of them are primes or a power of a prime.

Problem 8 I do believe :colone:

(edited 10 years ago)

Reply 645

We have totally messed up the numbers!

Solution 93

Let

F_{n}

be the

n

th Fibonacci number. From Catalan's identity, we have

F_{2n+1}^{2}+1=F_{2n-1}F_{2n+3}

, and

F_{2n-1}^{2}+1=F_{2n-3}F_{2n+1}

. Hence

F_{2n+1}^{2}+1 \equiv 0 \pmod {F_{2n-1}}

and

F_{2n-1}^{2}+1 \equiv 0 \pmod {F_{2n+1}}

Reply 646

Original post by Mladenov

Solution 93

Let

F_{n}

be the

n

th Fibonacci number. From Catalan's identity, we have

F_{2n+1}^{2}+1=F_{2n-1}F_{2n+3}

, and

F_{2n-1}^{2}+1=F_{2n-3}F_{2n+1}

. Hence

F_{2n+1}^{2}+1 \equiv 0 \pmod {F_{2n-1}}

and

F_{2n-1}^{2}+1 \equiv 0 \pmod {F_{2n+1}}

HAHAHAHAHAHAHAHAHAHA :lol:

absolute legend! :lol:

That took you less than a minute! :lol:

Reply 647

Original post by Mladenov

Problem 94**

One really nice functional equation.

Find all strictly increasing functions

f: \mathbb{R} \to \mathbb{R}

such that

f(x+f(y))=f(x+y)+1

, for all real numbers

x

and

y

x=0

, then

f(f(y))=f(y)+1

. Therefore, for all values of a in the range of f,

f(a)=a+1

This works as

f(x+f(y))=x+y+2=f(x+y)+1

and f is clearly a strictly increasing function

Does the question state that the range of f is all reals, or a subset of all reals?

Reply 648

Original post by Jkn

HAHAHAHAHAHAHAHAHAHA :lol:

absolute legend! :lol:

That took you less than a minute! :lol:

Number theory :tongue:

Original post by Zephyr1011

x=0

, then

f(f(y))=f(y)+1

. Therefore, for all values of a in the range of f,

f(a)=a+1

This works as

f(x+f(y))=x+y+2=f(x+y)+1

and f is clearly a strictly increasing function

Does the question state that the range of f is all reals, or a subset of all reals?

Yup, the range of

f

is the set of all real numbers.

(edited 10 years ago)

Reply 649

Original post by Mladenov

We have totally messed up the numbers!

Solution 93

Let

F_{n}

be the

n

th Fibonacci number. From Catalan's identity, we have

F_{2n+1}^{2}+1=F_{2n-1}F_{2n+3}

, and

F_{2n-1}^{2}+1=F_{2n-3}F_{2n+1}

. Hence

F_{2n+1}^{2}+1 \equiv 0 \pmod {F_{2n-1}}

and

F_{2n-1}^{2}+1 \equiv 0 \pmod {F_{2n+1}}

...That question took me 2 hours.

I mean, actually solving the question only took up around a quarter of that time, most of the rest was spent messing up the proof by induction, which I'd used to solve it and I've never heard of Catalan's Identity, but still...

Oh, and should I put the old Problem 93 back now, since there's a solution to it?

(edited 10 years ago)

Reply 650

Felix Felicis

Original post by Zephyr1011

x=0

, then

f(f(y))=f(y)+1

. Therefore, for all values of a in the range of f,

f(a)=a+1

This works as

f(x+f(y))=x+y+2=f(x+y)+1

and f is clearly a strictly increasing function

Does the question state that the range of f is all reals, or a subset of all reals?

On a side note, are you really 14? :shock:

Reply 651

Original post by Mladenov

Yup, the range of

f

is the set of all real numbers.

Great, then I think I've proved that

\forall a\in\mathbb{R}

f(a)=a+1

Reply 652

Original post by Felix Felicis

On a side note, are you really 14? :shock:

...Yes

Reply 653

Original post by Zephyr1011

Great, then I think I've proved that

\forall a\in\mathbb{R}

f(a)=a+1

You have proved that

f(x)=x+1

only on the curve

x=0

(not

f

my bad!), not over the whole complex plane

xOy

(edited 10 years ago)

Reply 654

Original post by Zephyr1011

...Yes

Mother of god... :O

Original post by Zephyr1011

Did you use the exact same method (i.e. fibionacci)? I may try to find an alternative solution (don't give away anything if you did use another method)

Reply 655

Original post by Mladenov

You have proved that

f(x)=x+1

only on the curve

x=0

(not

f

my bad!), not over the whole complex plane

xOy

What does

xOy

mean? And surely since any definition of f must be valid in the case

x=0

, and my definition is the only one valid in that case, my definition can be the only valid one.

Reply 656