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moneybird
Group Title
Denote by pk the kth prime number. Show that p1p2...pn +1 cannot be the perfect square of an integer.
 2 years ago
 2 years ago
moneybird Group Title
Denote by pk the kth prime number. Show that p1p2...pn +1 cannot be the perfect square of an integer.
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i am not sure i understand the question, but the product of n primes ( or the first n primes) cannot be a perfect square by the fundamental theorem of algebra. \[n=p_1^{\alpha _1}p_2^{\alpha _2}...p_j^{\alpha _j}\implies n^2=2p_1^{\alpha _1}p_2^{2\alpha _2}...p_j^{2\alpha _j}\]
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
You missed the plus one at the end of the product of primes
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
ooooh i see. sorry. i thought that was a subscript. so this number leaves a remainder of one when dividing by each smaller prime. i am sure you can use that to show it is not the square of something
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
As @satellite73 said, any number n can be written as:\[n=p_1^{\alpha _1}p_2^{\alpha _2}...p_j^{\alpha _j}\]\[\therefore n^2=p_1^{2\alpha _1}p_2^{2\alpha _2}...p_j^{2\alpha _j}\]and we know that:\[p_1p_2p_3...p_j+1\]is not divisible by any of the primes \(p_1\) to \(p_j\)therefore it cannot be the square of an integer? I /think/ this proves it?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
because the square HAS to divisible by primes less than itself and:\[p_1p_2...p_j+1\]is NOT divisible by any of the primes less than it.
 2 years ago

moneybird Group TitleBest ResponseYou've already chosen the best response.3
My proof is Suppose that \[p_{1}p_{2}p_{3}p_{4}p_{5}p_{6}...p_{j} + 1 = m^{2}\] \[p_{1}p_{2}p_{3}p_{4}p_{5}p_{6}..p_{j} = (m1)(m+1)\] m+1 or m1 must be an even number since 2 is a prime number m must be odd If m is odd, then m + 1 and m  1 are even numbers But there is only one even number in the set of prime numbers Therefore p1p2...pn +1 cannot be the perfect square of an integer.
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
how do you conclude the last part? But there is only one even number in the set of prime numbers < TRUE Therefore p1p2...pn +1 cannot be the perfect square of an integer. < WHY?
 2 years ago

asnaseer Group TitleBest ResponseYou've already chosen the best response.0
wait  I just got it!  your method is quite clever
 2 years ago
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