Curious Thing
Can you prove that a triangle with sides that can be written in the form n2+1, n2-1 and 2n ( where n>1) is right angled ?
Labels: mathschallenge
posted by Rajesh Lal at
7:08 AM
Wednesday, September 30, 2009
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9 Comments:
3, 4, 5 right triangle
n = 2
We have (n²-1)²+(2n)²=(n²+1)²
This satisfies Pythagoras's theorem with a triangle of sides
n²-1, 2n and hypoteneuse n²+1.
It can be used to generate a family of Pythagorean triples.
... that went down well :)
Here are the first 20 unique generated triples (after reduction):
{3,4,5}, {8,15,17}, {5,12,13}, {12,35,37}, {7,24,25}, {16,63,65}
{9,40,41}, {20,99,101}, {11,60,61}, {24,143,145}, {13,84,85}
{28,195,197}, {15,112,113}, {32,255,257}, {17,144,145}
{36,323,325}, {19,180,181},{40,399,401}, {21,220,221}, {44,483,485}
Just noticed {5,12,13} and {15,112,113} - sweet!
Chris, your proof simply the assertion rephrased. What about this:
Suppose n²+1 is not the hypotenuse of a right triangle with the catheti n²-1 and 2n.
Then (n²+1)² ≠ (n²-1)² + (2n)² (Pythagoras)
→ n⁴ + 2n² + 1 ≠ n⁴ - 2n² + 1 + 4n²
→ n⁴ + 2n² + 1 ≠ n⁴ + 2n² + 1
→ 0 ≠ 0, contradicion!
Ergo: n²+1 is the hypotenuse of a right triangle with the catheti n²-1 and 2n, q.e.d.
Oh, sorry, your proof was a little short, I didn't see how you got to "(n²-1)²+(2n)²=(n²+1)²" other than simply inserting n²-1, 2n and n²+1 into Pythagoras' theorem.
Hi mo. I thought my proof was the right length ;) The question was trivial; it didn't merit a detailed and rigourous proof.
I certainly didn't simply restate the question. I used the identity: (n²-1)²+(2n)²=(n²+1)² and observed that it fitted in with Pythagoras's theorem - not the other way round. I hadn't imagined that anyone wouldn't realise that (and possibly double-check it for themselves).
Hi mo. Sorry if the last was a bit terse, it wasn't meant to be.
Sometimes I give very short proofs. But when I'm more interested, watch out, I'll give reams of explanatory material. The more I write, the more I've enjoyed thinking about and crystallizing my understanding of the problem. Greetz.
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