<If p and q are integers p^2+3q^2 is an integer >1 if q not 0. Which means a+b>2c which is impossible (case 1)

since a+b is a positive integer and
a+b=2*c*(p^2 + 3*q^2) means 1/(p^2+3q^2) is an integer and divisor of 2*c. the smallest divisor is 2 thus (p^2 + 3*q^2)<(1/2)>

Do you agree, you were wrong again?

< al wazir: ...

Also,

a+b=2*c*(p^2 + 3*q^2)

is compatible with a + b < 2c.

a+b=2*c*(p^2 + 3*q^2)

is compatible with a + b < 2c.

THERE IS NO CONTRADICTION.>

I am not saying there is a contradiction since (1/2)<1. I am just saying a+b<2c and a+b=2*c*(p^2 + 3*q^2) implies p^2+3*q^2<1/2.

While you said:

<a+b<2c and a+b=2*c*(p^2 + 3*q^2) imply p^2+3*q^2<1, not p^2+3*q^2<1/2.>

(1/2)<p<1 and

a+b=2c((9/16)=9c/8<2c

Stop right there.

1/(p^2 + 3*q^2) = (a+b)/(2c).

1/(p^2 + 3*q^2) = (a+b)/(2c). >

Unfortunately for you

(a+b)/2c = p^2 + 3q^2

of which I never said it is an integer.

1/(p^2 + 3v^2) (the reciprocal of p^2 + 3q^2) however has to be an integer.

(a+b)/(2c) doesn't have to be an integer, so 1/(p^2 + 3*q^2) doesn't have to be either. In fact, if you're assuming that a+b < 2c, it can't be.>

I assume a+b<2c to show it cannot be. Indirect proof it is called among mathematicians. Do yourself a favour and read before you post, consult a second grader for proofreading and when ok then and only then post.

of which I never said it is an integer. >

Actually, <al wazir> is a consultant if I remember correctly. A consultant by definition is a person who knows 1000 ways to make love to a girl but cannot get a date.

By experience I can guess his approach when on a contract or negotiating one.

He will "show" how complex the job is, what sophisticated techniques are required and only he <al wazir> can deliver. That is cheating the customer by the way. Any similarities to his dancing and singing here are purely accidental.

What I remember from the Rogoff page receiving cheques even after the end of the contract I was under the impression you still are. Sorry.

You two get a room or something ..>

...or share a pi.>

A better approximation is 355/113 = 3.14159292... . But the second approximation isn't just closer to the exact value of pi, it's also *more efficient*. By that I mean that 22/7 uses three digits (2, 2, 7) to represent pi correctly to three significant figures, while 355/113 uses six digits (3, 5, 5, 1, 1, 3) to represent pi correctly to *seven* significant figures.

Is it possible to improve on that? Wikipedia (https://en.wikipedia.org/wiki/) lists several higher rational approximations, e.g., 52163/16604 and 103993/33102. But these are actually *less* efficient.

So I've written a script that does a brute-force search (the last recourse of the clueless). The closest rational approximation to pi with a numerator under 100,000 is 99733/31746 = 3.141592641592. The closest rational approximation with numerator under 1,000,000 is 833719/265381 = 3.141592653581. In terms of my definition of efficiency, neither is an improvement on 355/133.

If the numerator has N digits and the denominator has D digits, it seems to me that the bigger N and D are, the *less* likely such an expression is to represent pi to better than N+D significant figures. (But I can't prove it.)