## 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

1. satellite73

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. moneybird

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

3. satellite73

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

4. asnaseer

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?

5. asnaseer

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.

6. moneybird

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.

7. asnaseer

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?

8. asnaseer

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