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A math question

For a right-angle triangle with lengths of all three sides being integers, there must be at least one of the side of which its length is a multiple of 5. Prove this.

For example, a right-angle triangle can have sides with integral lengths:
3, 4, 5;
7,24,25... .
Each and every triple contains a multiple of 5.

People who do crptography should know the answer in 5 minutes. But the math needed for this question is just grade 6 level (I guess, I've never attended school in US. )
And this a theorem quite new that I don't think you can find it in textbooks.

Monday, October 7, 2002

The sides x, y, z of a right triangle are related as x2 + y2 = z2

Let's take a look at the last digit of squares...[looking at the multiplication table] Ok, it's one of (0,1,4,5,6,9).

Assumption: Let's assume lengths are not multiples of 5.
Then, the last digit of squared length is one of (1, 4, 6, 9).
There are only 6 combinations now [Checking all 6...] Yep, as I suspected. You can't find such integers A, B, C ending with one of [1, 4, 6, 9], that A + B = C.
Hence, our assumption is wrong and one of the legths is multiple of 5.

Yes, I suppose, 6th grade math was enough :)

P.S. In other words, modulo-10 addition operation can't be applied to the (1, 4, 6, 9) group.

Igor K.
Tuesday, October 8, 2002

Original Poster:  Correction to problem:
One and _only_ one of the 3 lengths is a multiple of 5.

Prove this.

Monday, October 28, 2002


15,20,25 - etc.

Tuesday, October 29, 2002

I assume the poster means "one and only one given no common factors for all three sides". This part is pretty easy. If the 3 sides have length 5a, 5b and c, then (5a)^2 +- (5b)^2 = c^2 (+- depending on which two sides are a multiple of 5).
This gives 25(a^2 +- b^2) = c^2.
This tells us that 25 is also a factor of c^2 and thus 5 is a factor of c.
To conclude, if 2 two of the sides have a factor of 5 then so does the third.

Paul Viney
Monday, November 4, 2002

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