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### Pythagorean Triples

```
Date: 07/14/97 at 08:09:27
From: Daniel Toscano
Subject: Pythagorean triples and extras

Is there a formula to determine the solutions to the following
equations?

a^2 + b^2 = c^2

(I have gained some info on this one by reading some of your
articles.)

Extending further:

a^3 + b^3 + c^3 = d^3

a^4 + b^4 + c^4 + d^4 = e^4

I am unable to work out solutions to these.  Can you help?

Note:  a^3 means - a to the power 3
b^4 means - b to the power 4
```

```
Date: 07/14/97 at 08:47:04
From: Doctor Jerry
Subject: Re: Pythagorean triples and extras

Hi Daniel,

From The Theory of Numbers, by Hardy and Wright:

"The most general solution of the equation x^2+y^2=z^2, satisfying
the conditions x>0, y>0, z>0, x and y have no common divisors, x is
divisible by 2 is x=2ab,  y=a^2-b^2,  and  z=a^2+b^2, where a, b
are integers of opposite parity and a and b have no common factors
and a,b>0. There is a 1-1 correspondence between different values
of a,b, and different values of x,y,z."

The Babylonians knew this theorem, more or less. Look at _The Exact
the Babylonians.

>a^3 + b^3 + c^3 = d^3
>
>a^4 + b^4 + c^4 + d^4 = e^4

I don't know the solutions to these equations. However, since it is
now known that equations of the form x^k+y^k=z^k have no integer
solutions for k>=3 (Fermat's Last Theorem), I would be surprised if
there were solutions.

-Doctor Jerry,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 07/25/97 at 07:58:05
From: Daniel Toscano

To Doctor Math:

Can you please find values for the unknown variables in the two
following equations:

w^3 + x^3 + y^3 = z^3

a^4 + b^4 + c^4 + d^4 = e^4

I know that it is possible for the variables to all have the value of
zero, but I would like to know if there are any other possible values
for the variables?  I am quite certain that for the second equation,
one variable must have a value of zero while the other variables take
on values in the thousands. I think the only way possible to discover
any solutions is through trial and error - but it would be interesting
to know if there is any mathematical way of discovering possible

Daniel
```

```
Date: 07/25/97 at 11:36:50
From: Doctor Rob
Subject: Re: Advancing from Pythagorean triplets

Several solutions are known for these equations.  For the first
equation the smallest are:

3^3 + 4^3 + 5^3 = 6^3
1^3 + 6^3 + 8^3 = 9^3
3^3 + 10^3 + 18^3 = 19^3
7^3 + 14^3 + 17^3 = 20^3
4^3 + 17^3 + 22^3 = 25^3
18^3 + 19^3 + 21^3 = 28^3
11^3 + 15^3 + 27^3 = 29^3

There are parametric formulae which give infinite sets of solutions,
but not necessarily all solutions.  See Albert H. Beiler, _Recreations
in the Theory of Numbers_, pp. 290-291.

Another reference is G. H. Hardy and E. M. Wright, _The Theory of
Numbers_, Section 13.7, pp. 199-201.

For the second equation, the smallest solution is:

30^4 + 120^4 + 272^4 + 315^4 = 353^4.

An even harder problem is to make a = 0, and Leonhard Euler
conjectured that this was impossible. He was wrong, and this was
proved possible by Noam Elkies, who found the first example. Roger
Frye found the smallest one:

95800^4 + 217519^4 + 414560^4 = 422481^4.

Again a reference is Hardy and Wright, Section 21.11, pp. 332-335.
No infinite families of solutions are known.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

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