Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    let [a,b] denote a pair of positive integers such that
    a^2 + b^2 = square number

    prove that if a is prime, there is a unique corresponding value for b.
    • 0 followers
    Offline

    ReputationRep:
    Offline

    0
    ReputationRep:
    (Original post by fishpaste)
    square numbers only have three factors, 1, their squareroot, and themself, apart from 1, which has 1. ( * )
    Bugger, just realised that isn't remotely true. Oh well. Sorry.

    However, it's only not true for squares of squares, perhaps you could adapt it or something.
    Offline

    15
    ReputationRep:
    If you instead break it down as

    a^2 = c^2 - b^2

    and use the fact that

    "a^2 has factors 1, a and a^2 only" - which is true as a is prime

    then you can uniquely work out b and c from a much as you did, remembering that a,b,c are positive.
    Offline

    0
    ReputationRep:
    (Original post by RichE)
    If you instead break it down as

    a^2 = c^2 - b^2

    and use the fact that

    "a^2 has factors 1, a and a^2 only" - which is true as a is prime

    then you can uniquely work out b and c from a much as you did, remembering that a,b,c are positive.
    a^2 = c^2 - b^2

    a^2 = (c + b)(c - b)

    since a is prime, a^2 has factors only 1,a,a^2

    the factors can't be (a)(a) because b is a positive integer, so they must be 1 and a^2 respectively

    so:
    c - b = 1

    this alone says that as long as you have c, you can find b, since b = c - 1
    QED
    (and a^2 = 2c - 1 = 2b + 1)

    like this?
    • Thread Starter
    Offline

    0
    ReputationRep:
    Well done, now try this extension:

    let [a,b] denote a pair of positive integers such that
    a^2 + b^2 = square number

    prove that there are as many different values for b as there are proper factors of a (Hence proving that prime numbers a have only one corresponding value for b, having only one proper factor)
    Offline

    0
    ReputationRep:
    (Original post by Tomp11)
    Well done, now try this extension:

    let [a,b] denote a pair of positive integers such that
    a^2 + b^2 = square number

    prove that there are as many different values for b as there are proper factors of a (Hence proving that prime numbers a have only one corresponding value for b, having only one proper factor)
    Same again, let the square number be c

    => a^2 = (c-b)(c+b)

    Now it follows that c-b can be 1,a or any proper factor of a, and so, since c is fixed, the number of possibilities for b is determined by the number of proper factors of a.
    Offline

    15
    ReputationRep:
    (Original post by beauford)
    Same again, let the square number be c

    => a^2 = (c-b)(c+b)

    Now it follows that c-b can be 1,a or any proper factor of a, and so, since c is fixed, the number of possibilities for b is determined by the number of proper factors of a.
    I'm not sure in what sense c is fixed; for example,

    8^2 + 15^2 = 17^2 and 8^2 + 6^2 = 10 ^2 [Two different cs]

    But more importantly, I don't believe the claim.

    I guess you have to count 1 (or a) as a proper factor (otherwise the original problem when a is prime is wrong). In which case 8 has three proper factors 1,2,4 but there are two solutions for b.
 
 
 
Turn on thread page Beta
Updated: August 22, 2004
Poll
Do you think parents should charge rent?
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.