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

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

Mathematics
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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. \[n=p_1^{\alpha _1}p_2^{\alpha _2}...p_j^{\alpha _j}\implies n^2=2p_1^{\alpha _1}p_2^{2\alpha _2}...p_j^{2\alpha _j}\]
You missed the plus one at the end of the product of primes
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

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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?
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.
My proof is Suppose that \[p_{1}p_{2}p_{3}p_{4}p_{5}p_{6}...p_{j} + 1 = m^{2}\] \[p_{1}p_{2}p_{3}p_{4}p_{5}p_{6}..p_{j} = (m-1)(m+1)\] m+1 or m-1 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.
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?
wait - I just got it! - your method is quite clever

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