The Proof is Trivial!

My intuition is that a very large class of numbers can be expressed in the forms he described as undesirable, especially if he allows a, b in Z. However I wonder if one can find a composite k such that it is not in one of the forms. Alas, I've ran out of procrastination time for tonight.

But I agree this isn't really heading in any useful direction, since there is no proof that the desirable solutions don't just generate, say, cycles of primes within a finite class.

I have not done an olympaid problem since last summer. Here's a geometric approach.

Solution 440

Let's look at the curve $y-x^{2}-nx-m =0$. We want to intersect the line $y=0$ twice in ${\mathbb{Z}}^{2}$.
To this end, we consider its discriminant - $\Delta = n^{2}-4m$. Naturally, we want to complete the square, so we choose $m = kn - k^{2}$, and thus we arrive at $\Delta = n^{2}-4kn+4k^{2} = (n-2k)^{2}$.
We now need a definition. A path consists of consecutive horizontal and vertical moves in a point lattice. In our case the points are at a unit distance.
We next turn to the concrete problem. Our initial curve is $y-x^{2}-x-2014$. The person who is supposed to have a winning strategy wants to get an equation with coefficients satisfying the above relation. In other words, we need a path starting from $(1,2014)$ and eventually, in particular after finitely many steps, intersecting a line of the form $y-kx-k^{2}=0$ in a lattice point.
It is clear that for $k=1$ there is no winning strategy (by symmetry the same holds for $k=-1$). For $|k| \not= 1$ every path $(1,2014) \mapsto (2,2014) \mapsto \cdots \mapsto (k_{n},l_{n-1})$ will eventually intersect this line since the gradient is not $\pm 1$ (this sequence of moves works when $k$ is positive).
Hobbes moves always away from the line since if after his move the path intersects the line we are done. So suppose that our path intersects the line after Calvin's move and the point of intersection is not a lattice point.
Our goal is to choose $k$ so that after Hobbes' move, Calvin wins or needs to move in order to win. Obviously, this is possible only when $|k|=2$. Indeed, let for example $k=2$; trivial algebraic computation shows that if Hobbes moves upwards, then either we are done, or we need a move to the right; the dual statement holds if he moves downwards.
Hence, if we set $k= \pm 2$, it is always possible to get a path intersecting the line in a lattice point.

I didnt say that Q has to be of any form other than the 4 forms i have stated clearly, if you substitute such a Q into the equation, with the only requiremnt of thos above, then we will yield twin prime solution.
that is what i have proved, but with added knowledge that there are an infinite number of Q not of the four forms we can find twin prime solution infinitely often. Btw please critisms of the work and not just vague things surrounding prime numbers. i appreaciate what you guys have been doing as i do believe we are simply fixing this soloutiona as it is certainly not irreperable.

...

so now we can consider ALL numbers of the form 6T+-1. these numbers can never be disivible by 6 and this is obvious.
now consider all numbers of the form
aS+-b. If S is of the form 6b+-1, then 6ab+-a+-b will be the expression that will arise as a result. But if S is not of the form 6b+-1 then a different expression will arise. Henec for all such values not of the form 6b+-1 we have a value of S that ensure that we not have an expression of the 6ab+-a+-b

Problem 440 *

Calvin and Hobbes are playing a game. They start with the equation $x^2+x+2014=0$ on the blackboard. Calvin and Hobbes play in turns, Calvin going first. Calvin's moves consist of either increasing or decreasing the coefficient of $x$ by $1$ and Hobbes' moves consist of similarly changing the constant term by $1$, with Calvin winning the moment an equation with integer solutions appears on the board. Prove that Calvin can always win.

[Source: Indian National Mathematical Olympiad 2014]

Luckily this solution can avoid latex (I hate doing latex on my phone).

Ramanujan says that the result is 'well known', so it shouldn't be too difficult.

Very few mathematical things were not well known to Ramanujan :P

Drying up?

Prove that

.

if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up

Are there any restrictions on the values of x1, x2 and n? Or can they take all real values?

Drying up?

Problem 442*

Prove that

.

Where $x_i$ are positive real numbers and $n$ is real.

if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up

Drying up?

Problem 442*

Prove that

.

Where $x_i$ are positive real numbers and $n$ is real.

if this isn't sort of familiar then opening this spoiler may be ruinous... if it is then open up

These are very short, but nice.

Problem 443**

$S$ an infinite collection of nested subsets of $\mathbb{N}$. Must $S$ be countable?

Problem 444**

Take a real sequence, let $Z$ be the set of its limit points. Must $Z$ be countable?

Can you define what you mean by nested subsets? The only definition I know of involves sequences of sets which are, of course, countable.

An almost as circular proof:
Since the inequality doesn't hold for n<1, and is clearly an equality for n=1, I will only consider n>1.
From Jensen's inequality:
$\displaystyle \sum_{n=1}^2 \frac 12 x^n \geq \left(\sum_{n=1}^2 \frac 12 x\right)^n$ (since x^n convex on $(0,\infty )$ )
The result then follows by simple algebra.

By the power-mean inequality...