The construction below shows that such tuples really hardly could be written in unique way. Maybe some exclusion from general case is due to the Hurwitz theorem about sum of squares and noncommutativity of quaternions. This construction is modification of Pietro Majer and Geoff Robinson suggestions.

Let us consider quaternion $q = t + x i + y j + z k$ and a constant quaternion $c$. I am not sure after reading of the question, if norm should be a square of some integer or not. It may be Hurwitz quaternion and simple nontrivial example is $c=(1+i+j+k)/2$. Now let us consider product $s = q c q$ as polynomial of $t, x, y, z$.

We have $|s| = |c| |q|^2 = |c| (t^2+x^2+y^2+z^2)^2 $, but $|s|=s_0^2+s_1^2+s_2^2+s_3^2$ for $s = s_0 + s_1 i + s_2 j +s_3 k$, where $s_0, s_1, s_2, s_3$ by definition are second order polynomial of $t, x, y, z$.

So different $c$ produces different tuples and $c=1$ produces solution mentioned by Pietro Majer and Geoff Robinson and in Mordell book.

The construction below shows that such tuples really hardly could be written in unique way. Maybe some exclusion from general case is due to the Hurwitz theorem about sum of squares and noncommutativity of quaternions. This construction is modification of Pietro Majer and Geoff Robinson suggestions.

Let us consider quaternion $q = t + x i + y j + z k$ and a constant quaternion $c$. I am not sure after reading of the question, if norm should be a square of some integer or not. It may be Hurwitz quaternion and simple nontrivial example is $c=(1+i+j+k)/2$. Now let us consider product $s = q c q$ as polynomial of $t, x, y, z$.

We have $|s| = |c| |q|^2 = |c| (t^2+x^2+y^2+z^2)^2 $, but $|s|=s_0^2+s_1^2+s_2^2+s_3^2$ for $s = s_0 + s_1 i + s_2 j +s_3 k$, where $s_0, s_1, s_2, s_3$ by definition are second order polynomial of $t, x, y, z$.

So different $c$ produces different tuples and $c=1$ produces solution mentioned by Pietro Majer and Geoff Robinson and in Mordell book.

[edit] To produce more general solution for second order polynomials it is possible to use some modification. Let's consider three constant quaternions $a, b, c$ with modules are squares of some integers. It may be done using Kac's method or something else. Then new solution is $a q c q b$.

The construction below shows that such tuples really hardly could be written in unique way. Maybe some exclusion from general case is due to the Hurwitz theorem about sum of squares and noncommutativity of quaternions. This construction is modification of Pietro Majer and Geoff Robinson suggestions.

Let us consider quaternion $q = t + x i + y j + z k$ and any constant quaternion $c$. I am not sure after reading of the question, if norm should be a square of some integer or not. It may be Hurwitz quaternion and simple nontrivial example is $c=(1+i+j+k)/2$. Now let us consider product $s = q c q$ as polynomial of $t, x, y, z$.

We have $|s| = |c| |q|^2 = |c| (t^2+x^2+y^2+z^2)^2 $, but $|s|=s_0^2+s_1^2+s_2^2+s_3^2$ for $s = s_0 + s_1 i + s_2 j +s_3 k$, where $s_0, s_1, s_2, s_3$ by definition are second order polynomial of $t, x, y, z$.

So different $c$ produces different tuples and $c=1$ produces solution mentioned by Pietro Majer and Geoff Robinson and in Mordell book.

The construction below shows that such tuples really hardly could be written in unique way. Maybe some exclusion from general case is due to the Hurwitz theorem about sum of squares and noncommutativity of quaternions. This construction is modification of Pietro Majer and Geoff Robinson suggestions.

Let us consider quaternion $q = t + x i + y j + z k$ and a constant quaternion $c$. I am not sure after reading of the question, if norm should be a square of some integer or not. It may be Hurwitz quaternion and simple nontrivial example is $c=(1+i+j+k)/2$. Now let us consider product $s = q c q$ as polynomial of $t, x, y, z$.

We have $|s| = |c| |q|^2 = |c| (t^2+x^2+y^2+z^2)^2 $, but $|s|=s_0^2+s_1^2+s_2^2+s_3^2$ for $s = s_0 + s_1 i + s_2 j +s_3 k$, where $s_0, s_1, s_2, s_3$ by definition are second order polynomial of $t, x, y, z$.

So different $c$ produces different tuples and $c=1$ produces solution mentioned by Pietro Majer and Geoff Robinson and in Mordell book.