Belated Happy Birthday, DX!

t MATH

I just know how to use some of it.

Good enough. Check my reasoning, okay? p and q are relatively prime. We have two numbers, p/4 and p^2 + 3q^2 (and we know all of these are integers). If p is not divisible by 3, then these numbers are relatively prime as well, because (this is the part I need to be checked on -- the above is true, I need to prove it) you can divide p^2+3q^2 by p/4 and get 4p + (3q^2 / (p/4)) (I'm kind of using the Euclidean algorithm a bit). Then the problem is reduced to 3q^2 and p/4 -- p/4 can't have any different factors from p, nor can q^2 from q, so the only way 3q^2 and p/4 can have any factors in common is if p is divisible by 3. Yes?

t / MATH Also, I've taken to writing M.Sh.L. at the end of proofs in my notes, just because it makes me think of you every time. Though I suppose, for completeness, I should figure out how to write it in Hebrew?

Nope, he can't possibly be. Rowlf first appeared on the Jimmy Dean Show (yes, the sausage guy used to be a singer) back in the early sixties, ten years before Waits's first album came out. Waits was only 14 when Rowlf started pounding away on his piano.

I am corrected. However, I am curious if Rowlf was always a piano playing dog, or if that aspect of his character developed later.

Breakfast This Morning: Sauteed mushrooms scrambled with eggs, chopped green onions, cheddar and gouda. Lattes & english muffins.

We're listening to Belle & Sebastian. JZ is reading my Mojo magazine about Kate Bush.

Matt should go to Stockholm. Other people should go to London. Maybe everybody should go to London.

Maybe everybody should go to London.

Yes, this.

::waits for everyone to start packing::

I would like to go to London! Although I should go someplace else first, as I've been to London. How's late May?

I vote for a F2F in Bruxelles. At the chocolate shop! ::totally giddy::

Then the problem is reduced to 3q^2 and p/4 -- p/4 can't have any different factors from p, nor can q^2 from q, so the only way 3q^2 and p/4 can have any factors in common is if p is divisible by 3. Yes?

Do you mean that since we get (p²+3q²)/(p/4)=4p+12q²/p and the first term, 4p, is obviously a multiplication of p, we have to show that the second term, 12q²/p is prime with p? Because if that's what you mean, then you show it quite clearly, IMNotMathyHO. The only thing that can connect between p/4 and 3q² is the 3, for exactly the reason you stated.

Though I suppose, for completeness, I should figure out how to write it in Hebrew?

I'm going to find a way to visually show you that, then. I'm thrilled that you prefer to use that!

Oh, if we're all going to meet in London, I can show you there, I guess.

I vote for a F2F in Bruxelles. At the chocolate shop! ::totally giddy::

We know a great chocolate shop in Philipsburg, St. Maarten.

IJS.

you show it quite clearly, IMNotMathyHO.

Yay! Then I managed to show that you can set up that product to be the product of coprime numbers even if p is a multiple of 3. So that's all good. But then... things get murky. I'm trying to work out part of Euler's proof that there's no integer solution to x^3 + y^3 = z^3, in case you were wondering. I know it's just elementary algebra and number theory, but it's totally beating me into the ground.

Math is hard.

If you're all going to be in London, will you come and visit me for lunch if I'm in Paris? My father's been promising to take me for about ten years now (since we saw Sabrina, I think).