Assuming that the triple $(a,b,c)=(2,3^{10}\cdot 109,23^5)$ found by Eric Reyssat is the one with the highest quality $q=\log(c)/\log(\text{rad}(abc))=1.6299\ldots$ for the ABC conjecture, one quickly computes that there is indeed no solution for $p_{n+1}=31$. Setting $Q=\text{rad}(KLp_{n+1})=\prod_{k=0}^{n+1}p_k$, we get $\log(L)\le q\log(Q)$. Running through the possibilities for these $L$ with prime factors $\le29$, one checks that $K=L-p_{n+1}$ either has prime factors $\ge31$, or $\prod_{k=0}^{n}p_k$ does not divide $K\cdot L$.

The (naive) pure python code which runs about 20 seconds on my machine for $p_{n+1}=31$ (which is $q$ in the third line) is

```
from math import prod, floor, factorial
q = 31
a = [z for z in range(2, q) if factorial(z-1)%z == z-1] # primes < q
P = prod(a)
qual = 1.63
c = floor((q*P)**qual) # upper bound for L
def La(K, a): # check if all prime divisors of K are in a
for p in a:
while K%p == 0:
K //= p
return K == 1
def indent(i): # Create iterator for candidates of L
return ' '*(4*i)
s = 'def tmp():\n'
s += ' c0 = c\n'
s += ' x0 = 1\n'
for i in range(len(a)):
s += indent(i+1) + f'b{i} = 1\n'
s += indent(i+1) + f'while b{i} <= c{i}:\n'
s += indent(i+2) + f'x{i+1} = x{i}*b{i}\n'
s += indent(i+2) + f'c{i+1} = c{i}//b{i}\n'
s += indent(i+2) + f'b{i} *= a[{i}]\n'
s += indent(len(a)+1) + f'yield x{len(a)}\n'
exec(s)
f = tmp()
for L in f:
K = L-q
if (K*L)%P == 0:
if La(K, a):
print(K, L)
break
else:
print('no solution')
```

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