Question

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

Answer 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.
\[n=p_1^{\alpha _1}p_2^{\alpha _2}...p_j^{\alpha _j}\implies n^2=2p_1^{\alpha _1}p_2^{2\alpha _2}...p_j^{2\alpha _j}\]

Answer 2

You missed the plus one at the end of the product of primes

Answer 3

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

Answer 4

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?

Answer 5

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.

Answer 6

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.

Answer 7

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?

Answer 8

wait - I just got it! - your method is quite clever