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Inequality problem

  1. Dec 17, 2013 #1
    Let [itex]a, b,[/itex] and [itex] c[/itex] be positive integers such that:

    [itex]\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}[/itex]

    Find the sum of all possible values of [itex]a[/itex] that are less than or equal to 100.


    My shot at it:
    I think may be that equation would also equals to a^2 + b^2 = c^2...
    So may be a is the sum of all the pythagorean triples in this particular case. But I'm not really sure and I'm a bit stuck!
    Please help
    Thanks a lot
     
    Last edited: Dec 17, 2013
  2. jcsd
  3. Dec 17, 2013 #2

    jedishrfu

    Staff: Mentor

    IS this the whole problem? There is no mention of b or c in the problem statement itself.

    My reading of it is that sum = 0 + 1 + 2 + 3 + 4 + ... + 100 is what they are looking for.

    Maybe the equation and the b and c values are used in a follow on problem.
     
  4. Dec 17, 2013 #3
    You can't use 0 as a since 1/0 is undefined. That must be excluded from your answer.
     
  5. Dec 17, 2013 #4
    It is not equivalent to [itex]a^2 + b^2 = c^2[/itex].
    For instance, take [itex]a = 3[/itex], [itex]b = 4[/itex].
    Then [itex]c^2 = \frac{1}{\frac{1}{9} + \frac{1}{16}} = \frac{144}{25} = \frac{12^2}{5^2}[/itex].

    Also, it's not all the natural numbers below [itex]100[/itex], as it needs to satisfy [itex]\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}[/itex], and [itex]a = 1[/itex] doesn't([itex]\frac{1}{1} + \frac{1}{b^2} > 1[/itex] which means that [itex]c < 1[/itex] which is not allowed).

    Rather, it would be equivalent to [itex]b^2c^2 + a^2c^2 = a^2b^2[/itex] or [itex]\frac{a^2 + b^2}{a^2b^2} = \frac{1}{c^2}[/itex](if you don't see this immediately, check). Can you give us the context in which this problem is given, as to get an idea what way the problem needs to be solved?
     
  6. Dec 17, 2013 #5
    Here's the whole problem, it's from a practice AMC 10 test #25.

    A49AE52.png

    I was able to solve 24 but the last one >.<
     
  7. Dec 17, 2013 #6
    I wrote a Java program and got 680. I know that is technically cheating, but hopefully I can offer some insight to an analytic solution (maybe).

    Some things I noticed:

    - If the ordered triple ##(a,b,c)## is a solution, the ordered triple ##(ka,kb,kc)##, k a non-zero integer (in general a complex number, but restricted to integer for this problem because we are working with integers), is also a solution because
    ##
    \begin{align*}
    \frac{1}{(ka)^2}+\frac{1}{(kb)^2} &= \frac{1}{(kc)^2}\\
    \frac{1}{k^2a^2}+\frac{1}{k^2b^2} &= \frac{1}{k^2c^2}\\
    \frac{1}{k^2}\left(\frac{1}{a^2}+\frac{1}{b^2}\right) &= \frac{1}{k^2}\frac{1}{c^2}\\
    \frac{1}{a^2}+\frac{1}{b^2} &= \frac{1}{c^2}
    \end{align*}
    ##

    - If ##(a,b,c)## is a solution, ##(b,a,c)## is a solution, which is trivial.
    - There were two "unique solutions." By the manipulations above, the one solution produced 11 solutions which totaled to 615 for the a terms being restricted to 100 or below. The other solution was a total anomaly with a solution triplet of (65,156,60). That gives a total of 680 for the a terms.


    I think the approach for finding the first triplet solution would be to work with the left side of the equation modified to: ##\frac{a^2+b^2}{a^2b^2} = \frac{1}{c^2}##. Work with the small primes 2,3, and 5 and try to end up with a perfect square in the denominator. Also note that ##9+16=25=5^2## which converts the sum of prime factor multiplication of 2 and 3 to 5s.

    I have no idea how to come up with the (65,156,60) solution working with just pen and paper.

    I imagine a lot of people answered 615 on the test.
     
    Last edited: Dec 17, 2013
  8. Dec 17, 2013 #7

    jbunniii

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    I don't have an analytical solution either, but here is an observation that might help to narrow the search.

    We can find an upper bound for ##b## in terms of ##a## as follows. First note that ##1/a^2 + 1/b^2 = 1/c^2## implies that ##1/c^2 > 1/a^2##, or equivalently ##c < a##. Since these are integers, this means that ##c \leq a-1##. Therefore,
    $$\frac{1}{a^2} + \frac{1}{b^2} \geq \frac{1}{(a-1)^2}$$
    which, assuming I didn't make an algebra mistake, rearranges to
    $$b^2 \leq \frac{a^2(a-1)^2}{2a-1}$$
    or
    $$b \leq \sqrt{\frac{a^2(a-1)^2}{2a-1}}$$
    This gives the following upper bounds.

    if a = 2 then b <= 1
    if a = 3 then b <= 2
    if a = 4 then b <= 4
    if a = 5 then b <= 6
    if a = 6 then b <= 9
    if a = 7 then b <= 11
    if a = 8 then b <= 14
    if a = 9 then b <= 17
    if a = 10 then b <= 20
    if a = 11 then b <= 24
    if a = 12 then b <= 27
    if a = 13 then b <= 31
    if a = 14 then b <= 35
    if a = 15 then b <= 38
    etc.
     
  9. Dec 17, 2013 #8

    Dick

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    Kudos to jbunnii and scurty for pushing on with this problem. Here's what I've got so far. If (a,b,c) is Pythagorean triple so a^2+b^2=c^2, then 1/(ca)^2+1/(cb)^2=1/(ab)^2, so the triple of integers (ca,cb,ab) solve the inverse problem. scurty's first solution comes from the first primitive Pythagorean triple, (3,4,5) and multiplying by 5. The second (anamoly) comes from the second primitive triple (5,12,13) and multiplying by 13.
     
  10. Dec 17, 2013 #9
    That is a very nice solution!
     
  11. Dec 17, 2013 #10

    Dick

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    Your work is impressive as well, I used it to help figure this out. But it's not quite a solution to the whole problem. It shows you how to generate solutions but it doesn't tell you those are ALL of the solutions. Should leave something for ThienNguyen to worry about.
     
  12. Dec 17, 2013 #11

    jbunniii

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    Nice!
     
  13. Dec 18, 2013 #12

    Dick

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    Thanks! But as I said scurty's numerical investigations showed the way. I wasn't smart enough to figure it out on my own.
     
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