Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta
    • Thread Starter
    Offline

    2
    ReputationRep:
    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
    • Community Assistant
    Offline

    13
    ReputationRep:
    Community Assistant
    (Original post by 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 (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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by 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.
    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
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by 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
    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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by 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.
    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:

    Suppose thatp1p2p3p4p5p6...pj+1=m2
    p1p2p3p4p5p6..pj=(m1)(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.
    • Study Helper
    Offline

    15
    Study Helper
    (Original post by 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:
    Supposing p_1...p_j+1=m^2, then

    p_1...p_j=m^2-1 = (m-1)(m+1).........(1)

    Since p_1=2 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.
    • Thread Starter
    Offline

    2
    ReputationRep:
    (Original post by ghostwalker)
    Supposing p_1...p_j+1=m^2, then

    p_1...p_j=m^2-1 = (m-1)(m+1).........(1)

    Since p_1=2 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.
    Thanks dude u always help me understand
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: February 1, 2015
Poll
Do you agree with the proposed ban on plastic straws and cotton buds?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.