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Search: Posts Made By: paulunderwood
Forum: Miscellaneous Math 2020-09-22, 17:02
Replies: 94
Views: 7,926
Posted By paulunderwood
Cool I can tighten things up by doing the following...

I can tighten things up by doing the following test (for any a=2^r):


gcd(a-2,n)==1
gcd(a-4,n)==1
kronecker(1-a,n)==-1
Mod(1-a,n)^((n-1)/2)==-1
Mod(2,n)^((n-1)/2)==kronecker(2,n)
...
Forum: GPU Computing 2020-09-22, 10:02
Replies: 7
Views: 183
Posted By paulunderwood
Here are parts of my /etc/default/grub ...

Here are parts of my /etc/default/grub
GRUB_CMDLINE_LINUX_DEFAULT="quiet amdgpu.ppfeaturemask=0xffffffff"
and /boot/grub/grub.cfg
linux /boot/vmlinuz-4.19.0-10-amd64...
Forum: GPU Computing 2020-09-22, 07:24
Replies: 7
Views: 183
Posted By paulunderwood
I ran apt-get update (Debian) and the system...

I ran apt-get update (Debian) and the system froze while installing ROCm-3.8.0 with the message "building initial module for 4.19-0-9-amd64". Rebooted. I had to do dpkg --configure -a (??) and it...
Forum: GPU Computing 2020-09-22, 04:50
Replies: 7
Views: 183
Posted By paulunderwood
Are you using Linux or Windows? Has the OS...

Are you using Linux or Windows? Has the OS upgraded recently?
Forum: Miscellaneous Math 2020-09-19, 17:44
Replies: 94
Views: 7,926
Posted By paulunderwood
Cool A variation on the theme

I have been considering x^2-2*x+2^r=0 and have been taking n+1 powers of x over it. I can also take n+1 powers over x^2-(4/2^r-2)+1 and expect the answer to be 1. This gives rise to (yet) another...
Forum: Miscellaneous Math 2020-09-19, 09:04
Replies: 94
Views: 7,926
Posted By paulunderwood
Letting r=1 is trivial as the Euler part becomes...

Letting r=1 is trivial as the Euler part becomes Mod(-1)^((n-1)/2). Also it is specified gcd(2-2^r,n)==1 because of x^2-(2^2/2^r-2)*x+1

r=2 does not pass gcd(4-2^r,n)==1. This comes from...
Forum: GpuOwl 2020-09-18, 09:56
Replies: 56
Views: 4,667
Posted By paulunderwood
I think you should have decreased it.

I think you should have decreased it.
Forum: Miscellaneous Math 2020-09-17, 16:41
Replies: 94
Views: 7,926
Posted By paulunderwood
Your idea seems right. I ran some of this: {...

Your idea seems right. I ran some of this:

{
forstep(n=30169231,30169231,2,
if(!ispseudoprime(n)&&!issquare(n),
for(b=1,n,if(gcd(b,n)==1,z=znorder(Mod(b,n));
for(r=1,z,A=Mod(b,n)^r;...
Forum: GpuOwl 2020-09-17, 05:24
Replies: 2,471
Views: 146,301
Posted By paulunderwood
I can't help. I have never used colab.

I can't help. I have never used colab.
Forum: GpuOwl 2020-09-17, 05:18
Replies: 2,471
Views: 146,301
Posted By paulunderwood
This is not a bash terminal. Are you using...

This is not a bash terminal. Are you using "colab"?
Forum: GpuOwl 2020-09-17, 05:05
Replies: 2,471
Views: 146,301
Posted By paulunderwood
No no. At the terminal enter g++ --version ...

No no. At the terminal enter g++ --version

Also try g++ followed by hitting tab a couple of times.
Forum: GpuOwl 2020-09-17, 04:43
Replies: 2,471
Views: 146,301
Posted By paulunderwood
What does g++ --version say? You might have...

What does g++ --version say?

You might have to hardwire your compiler name and version in gpuOwl's makefile.
Forum: Miscellaneous Math 2020-09-16, 13:42
Replies: 39
Views: 1,249
Posted By paulunderwood
Once again it is meaningless to talk about...

Once again it is meaningless to talk about "u>log(p)^2" when working with modular arithmetic. Also it makes absolutely no difference where you apply modulo operations. It is a "multiplicative...
Forum: Miscellaneous Math 2020-09-15, 18:52
Replies: 94
Views: 7,926
Posted By paulunderwood
Efficiencies

Using Euler+Frobenius on x=1+-sqrt(1-b^4). For intermediate values s*x+t in left-right exponentiation of the Frobenius test we have:-

Squaring:
? Mod(s*x+t,x^2-2*x+b^4)^2
Mod((2*s^2 + 2*t*s)*x +...
Forum: Miscellaneous Math 2020-09-15, 15:50
Replies: 12
Views: 299
Posted By paulunderwood
Mod(10,107)^10000000000000000000000000000000000000...

Mod(10,107)^100000000000000000000000000000000000000000
Mod(34, 107)

##
*** last result computed in 0 ms.

It actually works by left-right binary exponentiation modulo n...
Forum: Miscellaneous Math 2020-09-15, 15:21
Replies: 12
Views: 299
Posted By paulunderwood
This Robert at his computer: :primenet: ...

This Robert at his computer:
:primenet:

Know that a*b mod n is the same as (a mod n) * (b mod n) :smile:
Forum: Miscellaneous Math 2020-09-15, 14:06
Replies: 94
Views: 7,926
Posted By paulunderwood
I made a mistake: Case 4 of FLT...

I made a mistake: Case 4 of FLT (https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem#Proofs_for_specific_exponents) uses the fact a^4-b^4 is never a square. So I have to flip the sign to get:

...
Forum: Miscellaneous Math 2020-09-15, 10:44
Replies: 94
Views: 7,926
Posted By paulunderwood
test based on FLT proof n=4

In the previous I had x=1+-sqrt(1-2^r). I note that 2^r-1 (r>1) is never a square. So what else is never a square? In his proof of FLT case n=4 Fermat showed a^4+b^4 is never a square, which leads me...
Forum: Miscellaneous Math 2020-09-15, 09:06
Replies: 39
Views: 1,249
Posted By paulunderwood
:smile:

:smile:
Forum: Lounge 2020-09-14, 01:27
Replies: 5
Views: 310
Posted By paulunderwood
The B in IBM stands for Ben.

The B in IBM stands for Ben.
Forum: Miscellaneous Math 2020-09-13, 18:25
Replies: 9
Views: 371
Posted By paulunderwood
(2^29-2)%232 174 :ermm:

(2^29-2)%232
174


:ermm:
Forum: Miscellaneous Math 2020-09-13, 17:29
Replies: 39
Views: 1,249
Posted By paulunderwood
That might take some time However: ...

That might take some time

However:

{forstep(n=1001*1001,100000000,2,
if(!ispseudoprime(n),
for(u=1000000,n,
A=Mod(Mod([1+i,1;1,u],n),i^2+1);
X=A^n;...
Forum: Miscellaneous Math 2020-09-13, 17:05
Replies: 39
Views: 1,249
Posted By paulunderwood
{forstep(n=9,77,2, if(!ispseudoprime(n), ...

{forstep(n=9,77,2,
if(!ispseudoprime(n),
for(u=round(log(n)^2),n,
A=Mod(Mod([1+i,1;1,u],n),i^2+1);
X=A^n;
R=lift(lift(trace(X*i)))%i;
if(R==1||R==n-1,
print([n,u,R])))))}
[25, 11, 24]
[25,...
Forum: Miscellaneous Math 2020-09-12, 15:01
Replies: 39
Views: 1,249
Posted By paulunderwood
n=3225601;A=Mod(Mod([1+x,1;1,0],n),x^2+1);B=Mod(Mo...

n=3225601;A=Mod(Mod([1+x,1;1,0],n),x^2+1);B=Mod(Mod([1,1;1,x],n),x^2+1);R=lift(lift(A^n-B^n));print(R)
[x, 0; 0, 3225600*x]


Your "theorem" is wrong.
Forum: Miscellaneous Math 2020-09-11, 16:56
Replies: 39
Views: 1,249
Posted By paulunderwood
The characteristic equations of the matrices are:...

The characteristic equations of the matrices are:

x^2 - (1+u)*x + u-1 == 0 and
x^2 - (1+w)*x + w-1 == 0

The "residuals" depend on the jacobi symbol of the discriminants of these equations:
...
Showing results 1 to 25 of 1000

 
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