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Question Number 88207 by MAB last updated on 09/Apr/20

w h a t i s t h e b i g g e s t p r i m e p v e r i f y i n g :

p = \underset{k = 1}{\sum^{n}} [\sqrt{k}]

w h e r e n \in N and [x] i s f l o o r (x)

Commented by MJS last updated on 09/Apr/20

seems to be 197

I^{'} m working on an explanation

Answered by MJS last updated on 09/Apr/20

r = [\sqrt{k}]; k, q \in N

1 ⩽ k < 4 \Rightarrow r = 1

4 ⩽ k < 9 \Rightarrow r = 2

9 ⩽ k < 16 \Rightarrow r = 3

q^{2} ⩽ k < {(q + 1)}^{2} \Rightarrow r = q

we get q [2 q + 1] times

look at the sums p_{q} = \underset{k = 1}{\sum^{{(q + 1)}^{2} - 1}} [\sqrt{k}]

p_{1} = 3

now we start adding 2 s \Rightarrow uneven p^{'} s

p_{2} = 13

now adding 3 s \Rightarrow even & uneven p^{'} s

p_{3} = 34 = 2 \times 17

now adding 4 s \Rightarrow even p^{'} s

p_{4} = 70 = 2 \times 5 \times 7

now adding 5 s \Rightarrow 5 ∣ p

p_{5} = 125 = 5^{3}

now adding 6 s \Rightarrow even & uneven p^{'} s

p_{6} = 203 = 7 \times 29

now adding 7 s \Rightarrow 7 ∣ p

p_{7} = 308 = 2^{2} \times 7 \times 11

now adding 8 s \Rightarrow even p^{'} s

p_{8} = 444 = 2^{2} \times 3 \times 37

now adding 9 s \Rightarrow 3 ∣ p

p_{9} = 615 = 3 \times 5 \times 41

now adding 10 s \Rightarrow 5 ∣ p

p_{10} = 825 = 3 \times 5^{2} \times 11

now adding 11 s \Rightarrow 11 ∣ p

\dots

I {don}^{'} t have the time to continue

we must prove that \gcd (p_{q}, q + 1) \neq 1 \forall q > 6

Commented by MAB last updated on 18/May/20

y o u a r e i n t h e r i g h t w a y, c o n t i n u e .

h i n t : w h y {d o n}^{'} t y o u u s e v a l u a t i o n ?

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