I am aware that the sum of the reciprocal of the squares = π2/6 so I'd hazard the guess, yes, the required basis exists but I could be wrong.

In R2, an orthonormal vector v = <a1, a2> with a1 = 0 and a2 = any prime and vice versa has a 2-norm = sqrt(a12 + a22) that is the same as the chosen prime integer. I don't know what, "What about", refers to in the question though.

ETA: Just to add the Pythagorean triples like 3, 4, 5 - 5, 12, 13 etc... The 2-norm of a vector with coefficients a1=3 and a2=4 is the prime 5. &etc...