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

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satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1i 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. \[n=p_1^{\alpha _1}p_2^{\alpha _2}...p_j^{\alpha _j}\implies n^2=2p_1^{\alpha _1}p_2^{2\alpha _2}...p_j^{2\alpha _j}\]

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1ooooh 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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0As @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?

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0because 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.

moneybird
 3 years ago
Best ResponseYou've already chosen the best response.3My 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.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0how 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?

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