Primitives triangle with smallest side an even number

In summary, the conversation discusses the proof that if a right triangle has all sides rational and primitives, then one of the smaller sides must be an even number. Two approaches are presented, one using the Pythagorean theorem and the other based on the fact that the area of a right triangle with rational and primitive sides must be an integer.
  • #1
Adel Makram
635
15

Homework Statement


Prove that if a right triangle has all sides rational and primitives (co-primes), then one of the smaller side must be even number.

Homework Equations


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

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.
 
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  • #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.
 
  • #3
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.
 
  • #4
RUber said:
There is nothing in your second proof forcing d to be an integer. So I am not sure that is the best method.
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.
 
  • #5
Adel Makram said:
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.
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.
 
  • #6
Adel Makram said:
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.
There is no reason why d should be an integer, so no contradiction.
 
  • #7
Adel Makram said:
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.
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.
Adel Makram said:
But in the right triangle whose a, b are odd, c can not be even ( because c^2 is not divisible by 4).
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.
 
  • #8
RUber said:
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.
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.
 
  • #9
Adel Makram said:
2) If a and b are odds, c can not be even which makes the area of the triangle, d, not integer.
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.
Adel Makram said:
the area of the triangle, d, not integer which is a must for any right triangle.
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.
 

What is a primitive triangle?

A primitive triangle is a triangle with all sides being whole numbers and having no common factors other than 1. This means that the sides cannot be divided by any number other than 1 to result in another whole number.

What is the smallest side of a primitive triangle?

The smallest side of a primitive triangle is always an even number. This is because if the smallest side was an odd number, the other two sides would have to be odd as well, making the triangle not primitive.

How is the smallest side of a primitive triangle determined?

The smallest side of a primitive triangle can be determined by using the formula 2n+1, where n is any positive integer. This will always result in an odd number, which is then multiplied by 2 to get the smallest side of the triangle.

Can a primitive triangle have a side length of 0?

No, a primitive triangle cannot have a side length of 0. This is because 0 is not a whole number and it is also not a valid side length for a triangle.

Are there an infinite number of primitive triangles with an even smallest side?

Yes, there is an infinite number of primitive triangles with an even smallest side. This is because there are an infinite number of positive integers that can be used in the formula 2n+1 to get different side lengths for the smallest side.

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