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7:56 AM
Hello
$$(27a+5b)^2+(5a+2b)^2=(23a+5b)^2+(15a+2b)^2$$
$$(27a+3b)^2+(5a+2b)^2=(23a+2b)^2+(15a+3b)^2$$
 
8:29 AM
Generalised it :)
Let $P=2(d-b)$ and $Q=a^2+2ax+b^2$ and $R=c^2+2cx+d^2$
Then $$\{P(a+x)\}^2+(P\cdot b + Q - R)^2=\{P(c+x)\}^2+(P\cdot d + Q - R)^2$$
$\forall a,b,c,d,x$ :D
 
8:44 AM
Oh I forgot: this is implied by the fact that $a^2+b^2=c^2+d^2$
 
 
2 hours later…
10:46 AM
If $k$ is a square and $n+1$ is prime then $\left(\dfrac{m^2+1}{n+1}\right)=1$. Not sure what to do after this though.
Similarly $\left(\dfrac{k-1}{n+1}\right)=1$. Note that $m^2+1\not\equiv k-1\pmod{n+1}$.
If $n=k$ then $\left(\dfrac{-2}{k+1}\right)=1$ so $16\mid k$.
 
11:14 AM
@Haran @Mathphile any ideas
 
12:02 PM
Hello
@TheSimpliFire I have an idea on how to calculate pi by only using square numbers
(And no reimann zeta function!)
I'm aaking you because I don't know if it's a correct way of thinking, so I was hoping once I'd explain to you, you could hopefully see any errors. That would be much appreciated :)
Also @Mathphile and/or @RoddyMacPhee is it ok if at least one of you guys could code something: the total sum of the number of ways the integers from $0$ to say $N$ could be each written as the sum of two squares (they don't have to be distinct, so $0^2+1^2$ and $1^2+0^2$ counts as two ways)
I'm not sure if it's possible though xD sounds a bit specific
 
saving data. but for(n=0,N,forpart(n,s,forperm(...))
 
@RoddyMacPhee is what I am saying understandable? Sometimes im not great at explaining things
 
 
1 hour later…
I specifically said "And no reimann zeta function" ;)
 
Do you see $\zeta$ in my answer :P
 
xD
@TheSimpliFire how are you by the way :)
Why is it taking me five whole minutes to reply? :/
 
@MrPie chill
 
Its just my internet. Guess im so used to instantaneous happenings
 
1:28 PM
@MrPie cause your signals travel at 50656 m/s
 
Did you really just calculate that?
Ok so speed is distance over time, right?
Therefore you must be $253280$ metres away from my location.
So.... are you in Australia??? :P
 
 
1 hour later…
2:35 PM
@MrPie The UK is 9443 miles away from Australia :P
@MrPie ...and you multiply by 300 not 5
 
2:46 PM
Oh yes because there are 60 seconds per minute
XD
@TheSimpliFire thats probably the worst math mistake ive ever made
 
just blame it on physics
Anyway, what was your idea on the representation of pi
 
3:19 PM
@TheSimpliFire On what?
 
3:33 PM
@TheSimpliFire If the people would not have to pay for the climate change, they would not deny it. The matter is that we cannot influence the climate change. In particular, the CO2-claim is absurd. But not only that, modern comsology and the relativity theories are full of absurdities that are however ignored.
@Haran Something with pi ? I have no idea what he means.
 
4:32 PM
@Peter they get ignored because most other things it predicts have been tested to a degree.
he wanted to represent pi in relation to squares.
 
 
2 hours later…
6:44 PM
that's unlike the case of Flat earthers claiming gravity is about density where in the limit of a surface gravity it sort of is in newtonian physics $$GMm\over r^2 = GDVm\over r^2= G{4\over 3}Drm$$ if you take average density are radius of a globe. I'm not doing the convesion for flat earth.
forgot a factor of $\pi$ doh.
but this assumes being half buried.
 
 
2 hours later…
8:16 PM
@TheSimpliFire I still do not get the question about pi
 
That was asked to @MrPie
8 hours ago, by Mr Pie
@TheSimpliFire I have an idea on how to calculate pi by only using square numbers
which he hasn't replied yet
@Haran I'm still attempting the $n!+k$ problem. Earlier I thought of this approach using Wilson's theorem and I wondered if you had anything to add to it.
 
This prime pair is incredible :
0
A: $F_p,L_p$ both prime?

PeterIn fact, the numbers are related : $$L_n=F_{n-1}+F_{n+1}$$ The list of known prime numbers of both kinds reveals that we have a prime pair $(F_n,L_n)$ for the positive integers $$4\ 5\ 7\ 11\ 13\ 17\ 47$$ as well as the surprising pair for $n=148\ 091$ for which however $F_n$ is "only" a probable...

 
8:55 PM
hopefully my statement turns out to be useless.
Seems like a Goldbach elimination ... because one restatement of Goldbach is that forall m>3 there's a square you can subtract from $n^2$ that gives you a semiprime.
 
 
1 hour later…
10:21 PM
$m^2$ rather ...
 

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