The Proof is Trivial!

Original post by Lord of the Flies

Maybe one day, you'll agree on the beauty of this as well...

a,b\in\mathbb{N}

When

\dfrac{a^2+b^2}{1+ab}

an integer, it is a perfect square.

Spoiler

(by the way I am not posting this as a problem for the thread)

Took me a while to remember that incident :tongue:

Reply 1281

Zakee

Original post by Lord of the Flies

It is a */**, but a difficult one.

(the joke was that it requires a mathematical argument which I love and which bananarama despises - I'll let you figure out what it is from the picture)

Wasn't L'art's the gamma function, and yours was proof by infinite descent? :wink:

Reply 1282

theterminator

1

Has anyone got a solution for 192? :smile:

Reply 1283

Lord of the Flies

19

Original post by ukdragon37

Spoiler

Yeah but know what else isn't long enough? Your dissertation.

Original post by Zakee

Wasn't L'art's the gamma function, and yours was proof by infinite descent? :wink:

That sounds about right. Things have changed a bit though - I'm hitting on her twin sister now. :tongue:

Reply 1284

Zakee

Original post by Lord of the Flies

Yeah but know what else isn't long enough? Your dissertation.

That sounds about right. Things have changed a bit though - I'm hitting on her twin sister now. :tongue:

Who? Suzy? :wink:

. Way to go, alpha-male. Or should I say, beta-male. :tongue:

Reply 1285

ukdragon37

14

Original post by Lord of the Flies

Yeah but know what else isn't long enough? Your dissertation.

No worries, I'm making good progress :wink:

Spoiler

Reply 1286

bananarama2

Original post by ukdragon37

No worries, I'm making good progress :wink:

Spoiler

You definitely need iced coffee, not that coffee, in this weather ;P

Reply 1287

ukdragon37

14

Original post by bananarama2

You definitely need iced coffee, not that coffee, in this weather ;P

There's good weather outside? *checks* So there is.

Reply 1288

bananarama2

Original post by ukdragon37

There's good weather outside? *checks* So there is.

Reply 1289

Mladenov

1

Solution 185

Clearly,

f

is strictly increasing function; we set

\displaystyle f(-x) = f^{-1} (x)

. We note that

f(0)=1

and

f(-2x)= f^{-2}(x)

As

f > 0

over the whole real line, we can introduce

g(x) = \ln f(x)

, which is obviously increasing;

g(0) = 0

.
Now,

\displaystyle g(2x)+g(x) = \ln f(x) + \ln f^{2}(x) = 3g(x)

. Suppose

g(x) = xh(x)

. Thus

2xh(2x) + xh(x) = 3xh(x)

or

2xh(2x) = 2xh(x)

, for

x \not= 0

gives

h(2x) = h(x)

and hence

h(x) = h(0)

for all

x \in \mathbb{R}

.
Therefore,

f(x) = e^{h(0)x}

.

Solution 186

Imprimis, we notice that for each

m \in \{2,3,\cdots \}

, there exists

n \in \{1,2,\cdots, \}

such that

f(n)=m

.
Secondly,

f(n)

is injective. Otherwise,

m+1 = n+1

for

n \not= m

.
The equality

f(n) = 1

is impossible. If it were true, then we would have

1 = f(n) = f^{f(n)}(n) = n+1

, which is clearly not true.
Let

f(m)=2

,

m>2

. The inequality

m>2

holds true, for

f

does not have any fixed points!
Hence,

f^{f(1)}(1) = 2 = f(m)

or

f^{f(1)-1}(1) = m

. As

m>2

, we have

f^{f(1)-1}(1) = f^{f(m-1)}(m-1)

. It is obvious that

f(1)-1 > f(m-1)

, and thus

f^{f(1)-1-f(m-1)}(1)=m-1

. Following this path, we arrive at

f^{f(1)-1-f(m-1)-\cdots-f(2)}(1) = f^{f(1)}(1)

, which is a contradiction.

Solution 192

Suppose

b-a < \pi

. We can also suppose that

(a,b) \subset (0,\pi)

. Consider

\displaystyle g(x)=\frac{f(x)}{\sin x}

; let,

h(x)=\sin^{2}x g'(x)

. Then,

h'(x) \ge 0

over

(a,b)

. Moreover, there exists

c

such that

g'(c)=0

, which implies that

g(x)

decreases from zero to

g(c)

- contradiction.
Hence

b-a \ge \pi

.
Suppose

b-a= \pi

. On the one hand,

f(0)+f(\pi)=0

by the definition of

f

. On the other hand, however, we have

f(0)+f(\pi) >0

, from the condition

f+f'' >0

, which is a contradiction.

I suppose that

f

is continuous at

a

and

b

.

Solution 195

Let

a^{2},b^{2},c^{2}

be an arithmetic progression with non-zero common difference. Then,

a^{2}+c^{2}=2b^{2}

which has infinitely many solutions in positive integers. We derive this result from a more general theorem about ternary forms.

Solution 196

The elliptic curve

y^{2}=x(x-1)(x+3)

coincides with the modular curve

X_{0}(24)

. Hence

E(\mathbb{Q}) \cong \mathbb{Z}/{2\mathbb{Z}} \times \mathbb{Z}/{4\mathbb{Z}}

. Therefore,

[\pm1,\pm 1,\pm 1, \pm 1]

are the only

\mathbb{Q}

-rational points on the curve, and these points correspond to trivial solutions.

There are not enough number theory problems...

Problem 202**

Find all polynomials

P(x)

such that, for all

n \in \mathbb{Z^{+}}

, there exists at least one integer

m

such that

P(m)=2^{n}

.

Problem 203**

For any positive integer

n

set

A_{n} = \{j | 1 \le j \le n, \gcd(j,n)=1 \}

. Find all

n

such that

\displaystyle P(x) = \sum_{j \in A_{n}} x^{j-1}

is irreducible over

\mathbb{Z}[X]

.

Problem 204**

Let

p

be a prime number,

0 \le a_{1} < \cdots < a_{m}<p

and

0 \le b_{1} < \cdots < b_{n}

be arbitrary integers. Let

k

be the number of different reminders of

a_{j}+b_{i}

,

1 \le j \le m

and

1 \le i \le n

modulo

p

.
Prove that

m+n >p

implies

k=p

and

m+n \le p

-

k \ge m+n-1

.

(edited 10 years ago)

Reply 1290

such

5

Original post by Mladenov

...

I must say, I've never seen the word imprimis used before. It does add a certain credibility to your solution though :tongue:

I might try to sneak that word into my future solutions too!

Reply 1291

bananarama2

Original post by such

I must say, I've never seen the word imprimis used before. It does add a certain credibility to your solution though :tongue:

I might try to sneak that word into my future solutions too!

Lovely scrabble word, I should try and use it in the future.

Reply 1292

shamika

16

Original post by Lord of the Flies

Maybe one day, you'll agree on the beauty of this as well...

a,b\in\mathbb{N}

When

\dfrac{a^2+b^2}{1+ab}

an integer, it is a perfect square.

Spoiler

(by the way I am not posting this as a problem for the thread)

Prove it (it's supposed to be the hardest ever IMO question set, but I don't think that's true).

Reply 1293

Mladenov

1

Original post by shamika

Spoiler

The hardest IMO?! Vieta jumping kills it in three lines. For those who know not this method - click.

The difficulty of a given question is a relative concept. However, I would vote for IMO 2002 Problem 6.
In my view, the most difficult problem which has ever been proposed for IMO is Problem 6 Algebra IMO SL 2003.

Original post by such

Spoiler

Edgar Allan Poe is an influential writer, you know. :tongue:

(edited 10 years ago)

Reply 1294

Zakee

Original post by shamika

Prove it (it's supposed to be the hardest ever IMO question set, but I don't think that's true).

I tried to, today. (I've heard of Vieta Jumping but never bothered to look at it). Safe to say, my attempts have been futile. I've got somewhere, but all that is similar to a real solution is that I've used mathematics to yield more mathematics. :facepalm:

Edit: Looking at Vieta Jumping now, Mlad was right. Wow. That's so much simpler, haha.

Reply 1295

MW24595

Original post by Zakee

I tried to, today. (I've heard of Vieta Jumping but never bothered to look at it). Safe to say, my attempts have been futile. I've got somewhere, but all that is similar to a real solution is that I've used mathematics to yield more mathematics. :facepalm:

Edit: Looking at Vieta Jumping now, Mlad was right. Wow. That's so much simpler, haha.