# Pythagorean triples

• Jun 18th 2011, 10:40 AM
Mischa
Pythagorean triples
I have been trying to figure out this problem for hours with no luck.

There exist infinitely many primitive Pythagorean triples in which one of the legs is 8 (which means the difference between the other, longer, leg and hypotenuse is 8). Give the proof for this theorem and the three first triples, those with the shortest hypotenuse.
• Jun 18th 2011, 11:35 AM
Also sprach Zarathustra
Re: Pythagorean triples
Quote:

Originally Posted by Mischa
I have been trying to figure out this problem for hours with no luck.

There exist infinitely many primitive Pythagorean triples in which one of the legs is 8 (which means the difference between the other, longer, leg and hypotenuse is 8). Give the proof for this theorem and the three first triples, those with the shortest hypotenuse.

If x^2+y^2=z^2

2|x and gcd(x,y,z)=1

then:

x=2st, y=s^2-t^2, z=s^2+t^2
• Jun 18th 2011, 02:31 PM
tonio
Re: Pythagorean triples
Quote:

Originally Posted by Mischa
I have been trying to figure out this problem for hours with no luck.

There exist infinitely many primitive Pythagorean triples in which one of the legs is 8 (which means the difference between the other, longer, leg and hypotenuse is 8).

Which means, I'd say, that the difference between the squares of the hypotenuse and the other leg is \$\displaystyle 8^2=64\$

Tonio

Give the proof for this theorem and the three first triples, those with the shortest hypotenuse.

.
• Jun 18th 2011, 02:32 PM
tonio
Re: Pythagorean triples
Quote:

Originally Posted by Also sprach Zarathustra
If x^2+y^2=z^2

2|x and gcd(x,y,z)=1

then:

x=2st, y=s^2-t^2, z=s^2+t^2

How does this prove what the OP asked?

Tonio
• Jun 18th 2011, 02:44 PM
tonio
Re: Pythagorean triples
Quote:

Originally Posted by tonio
.

Suppose \$\displaystyle a^2+8^2=b^2 \iff 64=(b+a)(b-a)\Longrightarrow a+b<64\$ , and as we're talking about natural numbers there can only be

a finite number of such triples, so: what exactly did you try to prove?

Tonio
• Jun 19th 2011, 12:15 AM
Deveno
Re: Pythagorean triples
naturally i was curious, just how many such triples could we find? it was clear to me that b+a, being the larger of the 2 factors of 64, had to be 16 or 32.

b+a = 16 --> a = 6, b = 10. this is not primitive.

b+a = 32 --> a = 15, b = 17.

so it would appear there is but one such primitive triple: (8,15,17).

since the original question asks for the first 3 such triples, i suspect we are not answering the same question the OP is trying to ask.
• Jun 19th 2011, 11:29 AM
Mischa
Re: Pythagorean triples
x=2st, y=s^2-t^2, z=s^2+t^2

Is there other triples than 8,15,17? I could not find other with this theorem.
• Jun 20th 2011, 03:58 AM
tonio
Re: Pythagorean triples
Quote:

Originally Posted by Mischa
x=2st, y=s^2-t^2, z=s^2+t^2

Is there other triples than 8,15,17? I could not find other with this theorem.

DIdn't you read/understand what I and Deveno wrote? And what "theorem" are you talking about?

Tonio