Let $(p,q,r,s,t)$ be a quintuple of consecutive primes.
Are there infinite many such tuples such that $$p^2+q^2+r^2+s^2+t^2$$ is prime again ?
Motivation :
- $p^2$ is never prime
- $p^2+q^2$ is even for $p>2$
- $p^2+q^2+r^2$ is divisble by $3$ for $p>3$
- $p^2+q^2+r^2+s^2$ is even for $p>2$.
So, the smallest possible case is $5$ squares.
The following table, calculated by PARI/GP has the following meaning :
- The first column stands for the exponent $n$ in $10^n$,
- The second number is the number of primes below $10^n$
- The third column is the number of primes $p$ below $10^n$ such that the above property is satisfied.
? for(n=1,9,a=0;forprime(p=1,10^n,q=nextprime(p+1);r=nextprime(q+1);s=nextprime( r+1);t=nextprime(s+1);if(isprime(p^2+q^2+r^2+s^2+t^2)==1,a=a+1));print(n," ",p rimepi(10^n)," ",a)) 1 4 2 2 25 8 3 168 36 4 1229 258 5 9592 1637 6 78498 11129 7 664579 80041 8 5761455 603517 9 50847534 4706649 ?
A large example is $p=10^{100}+9631$
This looks like it may have definite crossover properties with a conjecture I floated a while back, that all primes of the form $x^2+1$ (for $x>90$) can be represented as the sum of five smaller such primes.
It seemed likely true in my case, so I would be surprised if your conjecture didn't turn out to be likely true as well.