Science and Technology Quiz 3

Status
Not open for further replies.
That is true! But you are missing an infinite number of other similar rules ;)

Edit: 13 = 2*2 + 3*3 but is not divisible by 3 or 9.
 
Edit: 13 = 2*2 + 3*3 but is not divisible by 3 or 9.
Mine was only an if divisible by 3 clause, it's nowhere near complete of course :)

That is true! But you are missing an infinite number of other similar rules
Countable though, I hope :p

/opens Excel again
 
Yup, countably infinite. They are a subset of the integers so, duh ;)

EDIT: Hah! Excel sucks at the infinite! Maths pwns physics!

EDIT2: I'm impressed so far, it shows you haven't cheated ;)
 
Well, let's pull a multiples of 5 that are a sum of squares out of Excel:
5-10-20-25-40-45-50-65-80-85-90-100-125-130-145-160-170-180-185
Let's divide by 5:
1-2-4-5-8-9-10-13-16-17-18-20-25-26-29-32-34-36-37
Now, that's funny, that's exactly the original sequence...

The same holds for 2 and for 13
 
That is true as well!

If x is a sum of two squares then 5x is also.

In fact, if a is a sum of 2 squares and b is too then so is ab.
 
Do I smell the term "prime factor" coming up?

n is a sum of two squares if and only if all it's prime factors are sums of squares.
That reduces the problem to finding out which prime factors are sums of squares. But there are infinitely many primes too :cry:

edit: damn, that's wrong, 9 is a sum of squares :(

edit 2: Restatement:
n is a sum of two squares if and only if either all it's prime factors are sums of squares or it's the square of an integer or twice the square of an integer.
 
Nope,

9 = 3*3 + 0*0

and 3 is not a sum of two squares.

You are very close though (look at your first reply).

EDIT: Drunk very early today ;)
 
edit 2: Restatement:
n is a sum of two squares if and only if either all it's prime factors are sums of squares or it's the square of an integer or twice the square of an integer.

I don't think so, it may take a while to come up with a counter-example however ;)
 
How about:
n is a sum of two squares if and only if all it's prime factors:
a) appear an even number of times in the factorization
b) are themselves the sum of squares

edit: x-post
 
Yes, that is correct. Now all you need to win is to state the condition for a prime to be the sum of 2 squares.

EDIT: Your statement is a bit wrong (EDIT: maybe, I've been drinking) since 5 is a sum of 2 squares and appears to an odd power in the factorisation.
But what I think you meant was:

n is a sum of 2 squares if and only if

1) n is a prime and a sum of 2 squares (need to elaborate to win)
2) n is composite and the primes in its factorisation which are not sums of 2 squares occur only to even powers.

EDIT2: Your statement is correct
 
Squaring an odd number will give an odd number.
Squaring an even number will give an even number.
Adding two odd or two even numbers will give an even number->not a prime (except 2)

So you need to add the square of an even number of our type to an odd number of our type squared. But that doesn't really help...

Well, that's it for maths for today, now it's time to watch television...
 
But you are so close ;)

All even numbers squared are of the form 2k*2k = 4k*k
Odd numbers are (2k+1)(2k+1) = 4k*k + 4k + 1
 
4k^2+4 l ^2+4 l +1=n
(n-1)/4=k^2+ l^2+ l

So n-1 needs to be divisible by 4. I'm not sure that's a sufficient condition though, it seems to be, according to my excel sheet. (IE. if all primes n with (n-1)/4 an integer are the sum of two squares)
(Advertisement breaks are good :))

edit: 133-1 is a 33*4, but 133 is not the sum of two squares. edit2: Oh, it's not a prime though :)
 
Nope!

2 = 1+1 ;)

But that is a rather unique prime ;)
 
Sooooooo....

n is a sum of two squares if and only if all it's prime factors:
a) appear an even number of times in the factorization
b) are of the form (j-1)/4
c) are 2
 
You are the winner!

I prefer to say if p is prime then it is a sum of 2 squares iff p/4 doesn't leave remainder 3.

Very well done!
 
My flatmate must have physically seen the past 10 posts or so flying about 3 inches over my head.
 
It's probably pi inches above
 
edit: Open floor! I've found a question.
 
Explain this Maths joke:

Q. What is an anagram of Banach-Tarski?
A. Banach-Tarski Banach-Tarski.
 
Status
Not open for further replies.
Top Bottom