Can you show that x² + y² = z² has many solutions in positive integers?

Question

We are familiar with 3² + 4² = 5², right?
But there is also 20² + 21² = 29².

Is there a general algorithm that would generate x, y and z as solution to x² + y² = z² where x, y and z are co-primes into each other?

Please help me I really wanted to know.

I'm very sorry David! I have to remind you that I want solutions in integers...

Hi David!

You're not quite right about multiplying x, y and z by a certain number "a". That would NOT make them co-primes into each other...

Answer 1

There are in fact infinite number of integer solutions to the equation x² + y² = z².

Proof:

Let z = x + (u/v)y where u, v are co=primes.

Thus vz = vx + uy.

Square both side to get

v²z² = v²x² + 2vxuy + u²y², z² = x² + y²

v²(x² + y²) = v²x² + v²y² = v²x² + 2vxuy + u²y²

Eliminate v²x² and rearrange

v²y² - u²y² = 2vxuy
(v² - u²)y² = 2vxuy

Divide both sides of the equation by y to get

(v² - u²)y = 2vxu

Here if put x = v² - u² and y = 2uv we have

z = x + (u/v)y = v² - u² + (u/v)(2uv) = v² + u²

THUS

x = v² - u²
y = 2uv
z = v² + u².

If uv is an even number and u and v are co-primes then we have an infinite solution to the equation x² + y² = z².

Done!

Answer 2

This type of equation is known as a Diophantine equation.

This particular one has solutions called Pythagorean triples.
The smaller relatively prime triples (x, y, z) are:

( 3, 4, 5)
( 5, 12, 13)
( 7, 24, 25)
( 8, 15, 17)
( 9, 40, 41)
(11, 60, 61)
(12, 35, 37)
(13, 84, 85)
(16, 63, 65)
(20, 21, 29)
(28, 45, 53)
(33, 56, 65)
(36, 77, 85)
(39, 80, 89)
(48, 55, 73)
(65, 72, 97)

65 = 5*13
72 = 2*2*2*3*3
97 = 97
They are relatively prime.

There exist some methods to generate pythagorean integer triples, but you must then check the values to see if they are relatively prime.

One method is called Euclid's formula.

x = a(2b+a)
y = 2b(a+b)
c = (a+b)^2 + b^2

Answer 3

the equation is true as long as x and y are sides of a right triangle, and z is the hypotenuse. 3-4-5 is just a popular one. multiply 3-4-5 times two and you get 6-8-10 (36+64=100). you can multiply it by 3, 4, 5, 6, 7, or any positive integer and it hold true. since there are infinite positive integers, there are infinite possible solutuions.

Now primes? that might be a little more difficult.


edited to add:


All the solutions I describe *are* integers! I'm saying that 3-4-5, 6-8-10, 9-12-15, 12-16-20 and so forth are all solutions consisting of positive integers. so basically, I am describing

(3a)^2+(4a)^2=(5a)^2 where a={1,2,3,4,5.... infinity} as an example of infinite solutions.

Answer 4

Given any prime number greater than 2, (Q, (Q²-1)/2, (Q²+1)/2) is a pythagorean triple.

Is it possible that the numbers aren't co prime? No. For a give Q, Q² is obviously a multiple of Q. Q²-1 and Q²+1 are NOT multuples of Q, so (Q²-1)/2 and (Q²+1)/2 are NOT multiples of Q.

Answer 5

There sure is.

You can generate ALL nonnegative solutions to x^2 + y^2 = z^2 by using

x = w(u^2 - v^2)
y = 2uvw
z = w(u^2 + v^2)

where u, v, and w are any nonnegative integers you like, provided u > v.

It is easy to see that a pair (x, y, z) of this form will work, because

[w(u^2 - v^2)]^2 + (2uvw)^2
= w^2(u^4 - 2u^2v^2 + v^4) + 4u^2v^2w^2
= u^4w^2 - 2u^2v^2w^2 + v^4w^2 + 4u^2v^2w^2
= u^4w^2 + 2u^2v^2w^2 + v^4w^2
= w^2(u^4 + 2u^2v^2 + v^4)
= w^2(u^2 + v^2)^2
= [w(u^2 + v^2)]^2.

It is harder to show that all possible solutions have this form, but they all do. If you are interested in discovering why, then I suggest you find a book on elementary number theory; it will almost certainly be proven in there.

I am not sure of an easy way to generate solutions in which x, y, and z are coprime. Sorry. (Even choosing u and v coprime and w = 1 is not good enough to guarantee your solution is coprime.)