@TobiasKildetoft, the OP asked about proving some huge number prime, which I thought was a waste of time in the first place. But I also claimed the task was impossible with current computers. A guy called TonyK gradually found evdence that the task was quite do-able and told me it was "time to recant" So I looked up the phrase "Mea Culpa" and typed i the whole prayer in Latin, using a poetry environment. I thought it was pretty good, and appropriate for the conversation that had taken place
@PeterTamaroff, the corrections robjohn mentioned give all you need. Note that the two constraints give an ellipse lying in a plane in $\mathbb R^3,$ so the extrema of distance (squared) from the origin are the semiminor and semimajor axes of that.
The quadratic form is homogeneous of degree $2$, the dot product is homogeneous of degree $1$ and the norm is homogeneous of degree $2$, so upon multiplication by $\bf x$, the equation $$0=\nabla f(x)+\lambda_1 \nabla h(x)+\lambda_2\nabla g(x)$$ becomes $$0=2f(x)+2\lambda_1 h(x)+\lambda_2 g(x)$$ But we have that $h(x)=xAx^t$ is $=1$ over the region and $g(x)=0$, so I get $$0=f(x)+\lambda_1$$, yet I have no restriction on $\lambda_2$. So I might as well chose it to be $0$?
I am using that if $f(x)$ is homogeneous of degree $p$; that is $f(ax)=a^pf(x)$, then $x\cdot \nabla f(x)=pf(x)$. @WillJagy
@Bageer You should suggest this to Mochizuki; instead of publishing it in traditional means, release a series of porn videos shot in the classroom with bits and pieces of his papers on the boards.
@Raindrop I was thinking about taking the putnam next semester, but I am not too sure (already taking too many classes+work+time to study what I want). Have you taken it before?(guessing you havn't by your questions). I think you should be asking the expert coach if that is a book you should study.
I think it requires some strange way of manipulating equations, but, for the tilme being, I cannot see how. Hope some smart one can come up with an ingenious solution.
@DanZimm I have heard that. I figured I would do it for the chance to do well (I hear if you do well you have a good chance in getting into any grad school) plus the teach who does this putnam prep class wants me to do it, although competition math isn't really my interest so I am still undecided. I think I will be looking at some past putnam problems, over the summer, to see if studying for the exam would even be fun for me.
We don't get to chose $x$ and $n$ we just know they exists from bezouts identity
We know some propertys of $x$ though
I think we should figure those out and see if they can be used, if not find another way to proceed I am to lazy to do either though
You can multiply both sides by $\gcd(a,b,c)$ or what ever sense it equals 1
this is prolly useless
We can get new solutions by
$(x+\frac{kb}{\gcd(b,c)},y-\frac{kb}{\gcd(b,c)})$
x+kb/gcd(b,c)=1?
does there exist a k so that that is true
k can be an arbitrary integer
i don't know and i am to lazy to do the work, this problem is probably very simple and I am just being stupid this happens to me in the morning, I am prolly going to have to go back and relearn a bunch of basic number theory now because i feel dumb
anyways, so we know $\gcd(\gcd(a,b),c) = 1$ and $\gcd(a,b) = v \neq 1$ so it follows that $v \mid b \iff \exists \; m : v = mb$ so then pick $n = v$ and it follows that $b + vc = b + mbc$ and now i lost my train of thpought
We know that $\gcd(a,\gcd(b,c))=1$. Let $b+c=(b,c)(P+Q)$. Note that $(P,Q)=1$. By Dirichlet's Theorem on primes in arithmetic progression there are infinitely many primes of the form $P \mod Q$ or in other words $\pi=P+nQ$. Obviously there is a $\pi$ such that $(a, \pi)=1$ and $\gcd(a,\gcd(b,c)\pi)=1$. Well $\gcd(b,c)\pi=\gcd(b,c)(P+nQ)=b+nc$. $\square$
I think people have a tendency to thumbs up things that look complicated but they can't or are to lazy to read, just because it looks like the op put alot of time in
