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Primitives triangle with smallest side an even number

  1. Nov 20, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that if a right triangle has all sides rational and primitives (co-primes), then one of the smaller side must be even number.

    2. Relevant equations
    For a right triangle (a,b,c) with c is the hypotenuse.
    $$a^2+b^2=c^2$$

    3. The attempt at a solution
    In order to create a contradiction, I assume both a and b are odd, so.
    $$a=2n_1 +1$$
    and.
    $$b=2n_2+1$$
    applying Pythagorean theorem,
    $$a^2+b^2=4(n_1^2+n_2^2-n_1-n_2)+2$$.
    This only gives me that c must be even but it does not tell me whether it is still rational and co-primes to a and b or not.
     
  2. jcsd
  3. Nov 20, 2015 #2
    Ok, I tried that, if a, b, c are co-primes, then the area of the triangle is:
    $$area=\frac{1}{2}ab=d$$
    $$ab=2d$$
    which means one of a or b must be even.
     
  4. Nov 20, 2015 #3

    RUber

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    There is nothing in your second proof forcing d to be an integer. So I am not sure that is the best method.
    Using your expansion from the Pythagorean theorem,
    ## a^2 + b^2 = c^2 = 4( n_1^2 + n_2^2 -n_1 - n_2 ) + 2 ##
    That gives that
    ##c = \sqrt{4( n_1^2 + n_2^2 -n_1 - n_2 ) + 2} = 2 \sqrt{ n_1^2 + n_2^2 -n_1 - n_2 +1/ 2} = 2\sqrt{K - 1/2} ##
    for some ##K \in \mathbb{Z}##.
    It should be pretty clear to show that c is not an integer.
     
  5. Nov 20, 2015 #4
    If both a&b are odd, d can not be integer. So, d can not be integer except when one of a&b is even which completes the proof without need for Pythagoras theorem.
     
  6. Nov 22, 2015 #5
    But this could be also applicable in any triangle, namely, the area of any triangle can not be a positive integer unless one of its sides is an even number. But in the right triangle whose a, b are odd, c can not be even ( because c2 is not divisible by 4). This means in order for the area of the right triangle to be a positive integer, a or b must be even.
     
  7. Nov 22, 2015 #6

    haruspex

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    There is no reason why d should be an integer, so no contradiction.
     
  8. Nov 23, 2015 #7

    RUber

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    What is d? d is the area. Nothing in your original problem said that the area had to be an integer...just that the lengths of all the sides were rational.
    I think your proof was supposed to show that if a and b are both odd, then c cannot be a rational number. Thus contradicting your original assumption and proving that if a, b, and c are all rational and primitives, either a or b must be even. You had the right attack using the Pythagorean thm.
     
  9. Nov 23, 2015 #8
    So, there are two attacks to proof it, both of them assume that a and b are odds:
    1) If a and b are odds, c can not be rational, contradicting the assumption that all are rationals and primitives.
    2) If a and b are odds, c can not be even which makes the area of the triangle, d, not integer which is a must for any right triangle. This means a or b must be even.
     
  10. Nov 23, 2015 #9

    haruspex

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    If a and b are odd, the area is not an integer. The reference to c being even is irrelevant, and it certainly does not by itself imply d is not an integer.
    You have no basis for that statement. Consider a=1, b=1. The area is not an integer, but you can construct a right triangle.
     
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