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2:20 AM
@OmegaKrypton do you like my latest code ?
 
what code?
 
see above 15:38
might speed up prime hunts ...
in fact a variant with restrictions required for one of @Peter 's recent questions shows n=17 with under 16000 classes of primes out of over 92000 mod 17 primorial would pass the fifth level.
next level brings it down to under 14000
must have messed up my conditions ... or his example is false. n!+p being prime for prime p. n=2 eliminates 1 mod 6, n=3 eliminates 29 mod 30, n=4 eliminates 11 mod 30, n=5 eliminates 6 mod 7, n=6 eliminates 1 mod 7, n=7 eliminates 2 mod 11, n=8 eliminates 5 mod 11 right ...
 
2:52 AM
for some reason I'm getting a null vector returned on that last condition... and I shouldn't be.
 
@RoddyMacPhee this?
 
my(r=[1],s=2);forprime(p=3,7,r=fold(concat,vector(p,i,apply(q->q+(i-1)*s,r)));s=s*p;r=select(w->gcd(w,s)==1,r));r
but with added conditions
doh local time was 15:38 lol GMT would have been 19:38 lol
 
3:07 AM
@OmegaKrypton have you figured out the code ?
 
3:27 AM
think I know what's wrong eliminating before the proper step due to cardinality not being checked ...
 
 
1 hour later…
4:34 AM
wait what language is this
 
 
6 hours later…
10:41 AM
PARI GP
 
 
1 hour later…
11:48 AM
6
Q: The equation $a^{4n}+b^{4n}+c^{4n}=2d^2$

HaranRecently, I found that if $a+b=c$, then $a^4+b^4+c^4=2d^2$ for some positive integer $d$. The parametric equation is: $$m^4+n^4+(m+n)^4=2(m^2+mn+n^2)^2$$ The condition $a+b=c$ (assuming $c \geqslant a,b$) isn't necessary. For example: $$7^4+7^4+12^4=2 \cdot 113^2$$ We can note that when we make t...

Bounty of +100 made available
 
12:09 PM
@Peter you might be interested
 
12:47 PM
@Haran Without doubt interesting, but how shall we arrive at the search limits you want ? We have to dramatically restrict the possible solutions or be content with much smaller search limits.
I think, if we cannot find an example with $d$ having , lets say , $100$ digits or less, I would consider this to be huge evidence.
 
1:14 PM
@Peter I have mentioned the required bounds in the comments
 
a,b,c are never congruent mod 3 ...
 
Can we classify all solutions of a^2+b^2+c^2 = 2d^2 ? This surely would help !
 
well if a^2 b^2 c^2 are the same mod 9, 6 needs to be a square mod 9
or they all need to be 0
same thing happens when all nonzero squares show up.
 
1:30 PM
@Peter Not sure...
 
There are only 64 combinations mod 9 and 9 of them just got eliminated ...
the RHS is always 0,2,8,5 mod 9
 
3
Q: Parametrization of $a^2+b^2+c^2=2d^2$

vuurIs a complete parametrization of primitive solutions to the equation $$a^2+b^2+c^2=2d^2\qquad a,b,c,d \in \mathbb{Z}$$ known? A reference would be great. Solutions to $a^2+b^2=c^2$ give solutions to the equation above, but I know that there are other solutions.

@Peter the above link has information on the complete parametrization of solutions of $$a^2+b^2+c^2=2d^2$$
The parametrization isn't that useful though. You can't easily work on checking all solutions in a bounded interval
 
The structure seems to be complex.
 
1:56 PM
@TheSimpliFire Hi
 
2:16 PM
@Peter updated my answer ...
probably an inaccuracy.
outter forvec for t,u inner forvec for v,w
 
2:57 PM
v,w must be at least 28.57 smaller that t,u for z not to be negative ...
@Peter I've went to n=36 and got 18506 under 10^7
 
3:22 PM
How far have you gone @Peter ...
` my(a=0);parforprime(p=1,10^7,[p%3,p%5,p%7,p%11,p%13,p%17,p%19,p%23,p%29,p%31,p%37],c,if(c[1]==2 &c[2]!=4 &c[2]!=1 & c[3]!=6 &c[3]!=1 &c[4]!=9 &c[4]!=6& c[4]!=10 &c[4]!=1 & c[5]!=12 &c[5]!=1 & c[6]!=14 &c[6]!=9 &c[6]!=16 &c[6]!=1 & c[7]!=18 &c[7]!=1 & c[8]!=19 &c[8]!=12 &c[8]!=22 &c[8]!=1 & c[9]!=7 &c[9]!=23 & c[9]!=24 &c[9]!=15 & c[9]!=28 &c[9]!=1 & c[10]!=30 &c[10]!=1 &c[11]!=4 & c[11]!=17& c[11]!=31 & c[11]!=19 & c[11]!=36 &c[11]!=1, print1(p",");a++));a` is my code so far.
 
3:54 PM
@TheSimpliFire sorry for being a pain.
 

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