Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    The question asks to prove that there are no solutions to the equation  a^3=b^3+3b^2+1 if  a and  b are both positive integers.
    Any ideas?
    Offline

    3
    ReputationRep:
    (Original post by Ano123)
    The question asks to prove that there are no solutions to the equation  a^3=b^3+3b^2+1 if  a and  b are both positive integers.
    Any ideas?
    This can be rewritten as a^3 = (b+1)^3 - 3b -> (b+1)^3 - a^3 = 3b -> (b+1-a)(b^2+2b+1+ab+a+a^2) = 3b. Maybe you can do something with that? I really have no idea.
    • Community Assistant
    • Welcome Squad
    Offline

    20
    ReputationRep:
    Community Assistant
    Welcome Squad
    (Original post by Ano123)
    The question asks to prove that there are no solutions to the equation  a^3=b^3+3b^2+1 if  a and  b are both positive integers.
    Any ideas?
    If b=a then obviously it doesn't hold as you would get 3b^2=-1 thus no real solutions.

    Say b>a - then a^3-b^3=3b^2+1 so it doesn't hold as the LHS is negative while RHS is positive.

    If b<a - then this becomes slightly trickier for a brain of mine at 2am. Try and figure it out from there.
    Offline

    15
    ReputationRep:
    Clearly  a>b, \forall a,b (real) as the RHS is obviously greater than  b^3 . We find values of b such that  a<b+1 .
    If  a<b+1 \Rightarrow a^3=b^3+3b^2+1<b^3+3b^2+3b+1 . Therefore  a<b+1 \text{ iff } b>0 . So if  b>0 then  b<a<b+1 . Of course no closed interval of length 1 contains three distinct integers. So any solutions if they do exist will have  b\leq 0.  \qed
    This is very similar to a STEP question,
    Offline

    17
    ReputationRep:
    (Original post by Ano123)
    The question asks to prove that there are no solutions to the equation  a^3=b^3+3b^2+1 if  a and  b are both positive integers.
    Any ideas?
    I have a super ugly proof I think

    So going from  \displaystyle a > b you can work through all the possibilities of signs a and b can have and find that a must be odd and b must be even (assuming both a and b are >0 of course).

    Then go back to  \displaystyle a^3 - b^3 = 3b^2 + 1 , realising that with  \displaystyle a > b the LHS doesnt have an upper "cap" eg you could set a is big as you like and b = 1 and get huge numbers so we should aim to make the LHS as small as possible, therefore we should set b to be the biggest integer possible, namely  \displaystyle b = a - 1

    doing this and subbing into  \displaystyle a^3 - b^3 = 3b^2 + 1  gives us;

     \displaystyle a^3 - (a-1)^3 = 3(a-1)^2 +1
    .
    .
    .
    solving this we get a= 1 and therefore b = 0 as a solution, however 0 isnt a positive integer so this solution is invalid.

    Common sense says that setting b = a - 3 will not work as the LHS will only get bigger and the RHS will get smaller meaning only a negative value of b will make LHS = RHS, this negative b may not even be an integer but we dont need to worry about it, since we must have b>0 there are no valid solutions.

    Edit: sat on the reply screen for an hour whilst I worked on paper, there already is an answer much better than mines, oh well
    Offline

    10
    ReputationRep:
    (Original post by Ano123)
    The question asks to prove that there are no solutions to the equation  a^3=b^3+3b^2+1 if  a and  b are both positive integers.
    Any ideas?
    let a = b+k (positive integers so a is clearly larger than b and k is a natural number greater than 0)
    (b+k)^3 = b^3 + 3b^2 +1
    The smallest number k can be is 1
    in this case
    (b+1)^3 = b^3 + 3b^2 + 1

    b^3 +3b^2 + 3b +1 = b^3 + 3b^2 +1
    3b = 0
    Already the left hand side is larger than the right hand side, any larger values of k will not work for the same reason.

    Therefore k is a natural number less than one - which does not exist - no k satisfies and therefore no a and b satisfy and therefore the equation has no solutions.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    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
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • 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

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