# Elementary Number theory question

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1)Let a, b1, b2 be non-zero integers, and suppose that a is relatively prime to both b1 and b2.

Prove that a is relatively prime to b1b2.

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

Can someone give hints how to start these

Prove that a is relatively prime to b1b2.

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

Can someone give hints how to start these

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1)Let a, b1, b2 be non-zero integers, and suppose that a is relatively prime to both b1 and b2.

Prove that a is relatively prime to b1b2.

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

Can someone give hints how to start these

**bigmindedone**)1)Let a, b1, b2 be non-zero integers, and suppose that a is relatively prime to both b1 and b2.

Prove that a is relatively prime to b1b2.

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

Can someone give hints how to start these

For (2) consider (p1p2 ...pk + 1) mod 4 and square numbers mod 4.

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For (1) consider what it means for a to be relatively prime to b1 and b2 in terms of their prime factors.

For (2) consider (p1p2 ...pk + 1) mod 4 and square numbers mod 4.

**SherlockHolmes**)For (1) consider what it means for a to be relatively prime to b1 and b2 in terms of their prime factors.

For (2) consider (p1p2 ...pk + 1) mod 4 and square numbers mod 4.

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Nothing is special about mod 4 because with mod 3 i get the same results that a square number leaves a remainder of 1 or 0

**bigmindedone**)Nothing is special about mod 4 because with mod 3 i get the same results that a square number leaves a remainder of 1 or 0

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With mod 3, p_2=3, and so (p1p2 ...pk + 1) = 1(mod 3) - so you can't use the fact that a square =0 or =1 (mod3) to prove that it can't be a square.

**ghostwalker**)With mod 3, p_2=3, and so (p1p2 ...pk + 1) = 1(mod 3) - so you can't use the fact that a square =0 or =1 (mod3) to prove that it can't be a square.

Suppose thatp1p2p3p4p5p6...pj+1=m2

p1p2p3p4p5p6..pj=(m−1)(m+1)

m+1 or m-1 must be an even number since 2 is a prime numberm must be oddIf m is odd, then m + 1 and m - 1 are even numbersBut there is only one even number in the set of prime numbersTherefore p1p2...pn +1 cannot be the perfect square of an integer.

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I see a solution online for the question but i dont understand it can you explain this a bit more to me although i am not sure im suppose to use this method:

**bigmindedone**)I see a solution online for the question but i dont understand it can you explain this a bit more to me although i am not sure im suppose to use this method:

.........(1)

Since being the first prime, then LHS in (1) is even.

Hence, m-1 or m+1 must be even.

If one is, so is the other, and so they are both even.

So the RHS as at least two factors of 2.

But the LHS has only one factor of 2, the first prime, the rest being odd.

Hence the two cannot be equal and there can be no m satisfying the original equation.

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(Original post by

Supposing , then

.........(1)

Since being the first prime, then LHS in (1) is even.

Hence, m-1 or m+1 must be even.

If one is, so is the other, and so they are both even.

So the RHS as at least two factors of 2.

But the LHS has only one factor of 2, the first prime, the rest being odd.

Hence the two cannot be equal and there can be no m satisfying the original equation.

**ghostwalker**)Supposing , then

.........(1)

Since being the first prime, then LHS in (1) is even.

Hence, m-1 or m+1 must be even.

If one is, so is the other, and so they are both even.

So the RHS as at least two factors of 2.

But the LHS has only one factor of 2, the first prime, the rest being odd.

Hence the two cannot be equal and there can be no m satisfying the original equation.

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