Pierino and the Riddle of the Three Squares

One infinite set of solutions includes $x = n$ , $y = 4n$ , $z = 8n$ , and $t = 9n$ for any positive integer $n$ , since $n^2 + (4n)^2 + (8n)^2 = (9n)^2$ .

Let’s consider the equation

$t^{2} = x^{2} + y^{2} + z^{2}$ with t , x , y , z $\in$ N and x , y , z different.

We can consider

t = a + b

Then

$(a+b)^{2} = x^{2} + y^{2} + z^{2} = a^{2} + b^{2} + 2ab$

Therefore we can write that

$x = a$

$y = b$

$z^{2} = 2ab$

The three squares must have different sides, therefore

$x^{2} \neq z^{2} \Rightarrow a^{2} \neq 2ab \Rightarrow a \neq 2b$

$y^{2} \neq z^{2} \Rightarrow b^{2} \neq 2ab \Rightarrow b \neq 2a$

$x \neq y \Rightarrow a \neq b$

$z^{2}$ must be the perfect square of an integer number z , but it must also be the product of three factors ( 2ab ), therefore we can think of writing it as

$z^{2} = p^{2} q^{2}$

We can now arbitrarily assign factors a and b to p and q:

$2a = p^{2}$ (1)

$b = q^{2}$ (1)

In this way, if a solution exists, it will have the shape that was imposed. If we do not find a solution, that will NOT be enough to say that no solution exists at all, but only that no solution exists of that shape. On the contrary, if we find a solution, that solution will have the shape given by the previous assumptions.

From (1) we can say that:

• a contains an odd number of factors “2”

• b is either an odd number or it contains an even number of factors “2”

Therefore, we can factorize a and b as:

$a = 2a_1a_2a_3 ... a_n 4^f$

$b = b_1b_2b_3 ... b_m 4^g$

where f , g $\in$ N

Thus:

$2ab = 2 \prod_{i=1}^{n} 2a_{i}4^{f} \prod_{j=1}^{m} b_{j}4^{g} = z^{2}$

$2ab = 4^{f+g+1} \prod_{i = 1}^{n_1} a_i^{2h_i} \prod_{j = 1}^{n_2} b_j^{2h_j} \prod_{r_1 = 1}^{s_1} a_{r1}^{2h_{r1}+1} \prod_{r_2 = 1}^{s_2} b_{r2}^{2h_{r2}+1} = z^{2}$

with $n_1 < n$ , $n_2 < m$ .

The first two products correspond to the factors raised to an even power, while the latter two products correspond to the factors raised to an odd power.

If the expression

$2ab = 4^{f+g+1} \prod_{i = 1}^{n_1} a_i^{2h_i} \prod_{j = 1}^{n_2} b_j^{2h_j} \prod_{r_1 = 1}^{s_1} a_{r1}^{2h_{r1}+1} \prod_{r_2 = 1}^{s_2} b_{r2}^{2h_{r2}+1}$

has to be equal to a perfect square $z^{2}$ , then $s_1 = s_2 = s$ , $r_1 = r_2 = r$ , $a_r = b_r \forall r = 1 ... s$

We have obtained a description, in terms of factorization, of a and b :

• For both a AND b , odd factors should be raised to an even power;

• For both a AND b , if an odd factor is raised to an odd power, then the same factor must be present in the other number ( b or a ), raised to an odd power, so that the sum of the two powers is an even number;

• For a OR b , factor “2” must be raised to an odd power, and for the other number ( b or a ) factor “2” must be raised to an even power

Therefore, there are infinite sets of numbers t , x , y , z $\in$ N and x , y , z different, that solve the equation

$t^{2} = x^{2} + y^{2} + z^{2}$

where

$t = a + b$

$x = a$

$y = b$

$z^{2} = 2ab$

and where a and b can be represented as follows:

$a = 2 * 4^{f} \prod_{i = 1}^{n_1} a_i^{2h_i} \prod_{r = 1}^{s} a_{r}^{2h_{ra}+1}$

$b = 4^{g} \prod_{j = 1}^{n_2} b_j^{2h_j} \prod_{r = 1}^{s} b_{r}^{2h_{rb}+1}$

where

$a_r = b_r \forall r = 1 ... s$

$h_{ra}, h_{rb} \in N, \forall r = 1 ... s$

Examples:

$a = 2, b = 9 \Rightarrow 2^2+9^2+2*2*9 = 4 + 81 + 36 = 121 = 11^2 = 2^2+9^2+6^2$

$a = 2*3^2=18, b = 5^2=25 \Rightarrow 18^2+25^2+2*18*25 = 324 + 625 + 900 = 1849 = 43^2 = 18^2+25^2+30^2$

One last note: $a = 2, b = 1$ is not a solution because

$a \neq 2b$

$b \neq 2a$

$a \neq b$

Let $(a,b,c)$ be a primitive Pythagorean triple (so $a^2 + b^2 = c^2$ ). Set $x=a^2$ , $y=ab$ and $z=bc$ . Then $\begin{aligned} x^2 + y^2 + z^2 &= (a^2)^2 + (ab)^2 + (bc)^2 \\ &= a^2(a^2 + b^2) + b^2c^2 \\ &= a^2c^2 + b^2c^2 \\ &=c^2(a^2+b^2) \\ &= (c^2)^2 \end{aligned}$ Since $(a,b,c)$ is primitive, $a^2$ , $ab$ and $bc$ are distinct. Thus, $t=c^2$ is a solution when $c$ is the hypotenuse of any primitive Pythagorean triple.

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As an example, taking the primitive Pythagorean triple $3,4,5$ yields $(x,y,z,t) = (9, 12, 20, 25)$ . It is indeed true that $\begin{aligned} 9^2 + 12^2 + 20^2 &= 25^2 \\ 81 + 144 + 400 &= 625 \\ 625&=625 \end{aligned}$

For other ways of generating $(x,y,z,t)$ , read about Pythagorean quadruples .