Find all solutions in integers to a^2 + b^2 + c^2 = d^2, where a, b and d are prime.

I assume that the condition for the number c is to be an integer,
otherwise the question is trivial.  Also, I assume all numbers to be
>= 0, which doesn't change the nature of the problem.

Let's use a little modulo arithmetic.  The following two tables will be
of use:

x (mod 4)   x^2 (mod 4)        x (mod 3)   x^2 (mod 3)
-----------------------        -----------------------
0           0                  0           0
1           1                  1           1
2           0                  2           1
3           1

The notation {0,1} used below is used to indicate that either 0 or 1
is possible here.

Start of solution:

d can't be 2, since then a^2+b^2+c^2 = 4, which is impossible under
the given conditions.  Therefore, d, being a prime, must be odd (and
therefore = either 1 or 3 modulo 4).

Applying modulo 4 arithmetic on the original equation,

 a^2  +  b^2  +  c^2  = d^2

{0,1} + {0,1} + {0,1} = 1       (modulo 4),

which shows that exactly two of a,b,c must be even.  Since at least one
of them is a prime, that prime is 2.  Let's say b = 2.

The new equation is taken modulo 3,

 a^2  + 4 +  c^2  = d^2

{0,1} + 1 + {0,1} = {0,1}       (modulo 3),

which gives the two possible patterns

1 + 1 + 1 = 0           and             0 + 1 + 0 = 1.

Looking at the former, it says that d^2 is divisible by 3, which can
only mean that d = 3, since d is a prime.  So, the equation now reduces
to a^2 + c^2 = 5, which has the single solution a = 2, c = 1, since
only c is permitted not to be a prime.

The other pattern indicates that a^2 is divisible by 3, which leads to
a = 3, and the equation is reduced to 13 + c^2 = d^2, or
13 = (d-c)(d+c).  Since 13 is a prime, it follows that d-c = 1 and
d+c = 13, which gives d = 7 and c = 6.

Thus, the solutions are

a = 3, b = 2, c = 6, d = 7
a = 2, b = 2, c = 1, d = 3,

as well as

a = 2, b = 3, c = 6, d = 7,  being a permutation of the first.

-- Arndt Jonasson (posted by Chris Long)

As a footnote, the general solution to

        a^2 + b^2 + c^2 = d^2                                    (1)

has been known for quite a while.  Robert Spira, "The Diophantine
Equation x^2 + y^2 + z^2 = m^2", American Mathematical Monthly, vol 69,
1962 pages 360-365 provides the following parametric solution:

    a = 2(rs+tu)   b = 2(ru-st)   c = r^2+t^2-s^2-u^2
    d = r^2+s^2+t^2+u^2                                          (2)

He has a list of references also.  It is a harder problem to find
solutions of (1) where the sum of any two a,b,or c squared is a square,
especially when all 3 sums are required to be squares.  See Problem
D18 in Richard K. Guy's "Unsolved Problems in Number Theory" 1st ed.
Springer-Verlag, New York.

-- Randall L. Rathbun <Randall.Rathbun@SanDiego.NCR.COM>