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Is it Possible to Prove that...

Date: 10/28/97 at 20:21:02
From: Marlon Coates
Subject: Can you prove that...

If the hypotenuse of a right angle triangle is divisible by 4, can you 
prove that the legs are also divisible by 4?

Date: 10/30/97 at 08:49:00
From: Doctor Bruce
Subject: Re: Can you prove that...


I believe you are asking this question: If the length of each side of 
a right triangle is a whole number, and the length of the hypotenuse 
is divisible by 4, then is the length of each leg also divisible by 4?

Yes, this is true.  We begin with the Pythagorean Theorem, which 

     a^2 + b^2  =  c^2

where c is the hypotenuse and a,b are the legs of a right triangle.  
We are assuming now that c is divisible by 4, so certainly c is an 
even number and therefore c^2  is divisible by 4 (in fact, c^2 is 
divisible by 16, but we don't need to use that information just yet).

Now, a^2 and b^2 are two numbers that add up to an even number, so 
a^2 and b^2 are both odd numbers or else they are both even numbers.  
We will show that it is impossible for them both to be odd numbers.

When we square an odd number, we always get a number which leaves a
remainder of 1 when we divide it by 4. For example,

     3^2 =  9;   9 divided by 4 =  2 with remainder 1
     5^2 = 25;  25 divided by 4 =  6 with remainder 1
     7^2 = 49;  49 divided by 4 = 12 with remainder 1

and so on. You might try to show that this is true for every odd 

Then, if we square two odd numbers and add them together, the result
will leave a remainder of 2 when divided by 4. So we are saying if
a and b are odd, then a^2 + b^2 has remainder 2 when divided by 4.
But a^2 + b^2 is equal to c^2, which has no remainder when divided 
by 4. So, the case of a and b both being odd can never happen.

So a and b must both be even numbers, when c is  divisible by 4. 
We can just divide the Pythagorean Theorem through by 4 and get

     (a/2)^2 + (b/2)^2 = (c/2)^2.

This means, we know that (a/2), (b/2), and (c/2) are all whole 
numbers. But we know more - (c/2) is actually an even number, since  
c itself was divisible by 4. So now we can apply all the same logic to 
the new numbers (a/2), (b/2), and (c/2) to deduce that (a/2) and (b/2)  
are both even numbers. That means both a and b were divisible by 4 to 
begin with.

Hope this helps,

-Doctor Bruce,  The Math Forum
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Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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