# About sum of n squares

The theorems about sum of two, three and four squares are very well known, but, i am just asking out of curiosity, is there any kind of algorithm or formula to determine whether a number is a sum of $$n$$ squares? I mean for a specific $$n$$, is there some relation between that algorithm and the sum? Edit: I understand i was not precise.... I mean leaving 0s and not specifically 2 or 3 squares.

• Do you mean consecutive squares or just random squares? – Javi Apr 8 at 14:05
• Random squares sir – user662075 Apr 8 at 14:06

Once you know that any number is the sum of $$4$$ squares you know that it is the sum of $$n$$ squares for $$n \gt 4$$ because you can add as many copies of $$0^2$$ as you need.

Added: if you want to express $$N$$ as the sum of $$n$$ squares without using $$0$$ the big problem comes if $$N$$ is too small. Clearly you need $$N \ge n$$. The natural thing to do is start with $$n-4\ 1$$'s and express $$N-n+4$$ as a sum of four squares Because you cannot express $$1,2,3,5,6,8,9 \ldots$$ as a sum of four nonzero squares you cannot express $$n+1,n+2,n+3,\ldots$$ as a sum of $$n$$ squares. I did not find this sequence in OEIS. This answer shows that every natural greater than $$34$$ is the sum of five nonzero squares. We can then say that for $$n \gt 4,$$ all numbers greater than $$n+29$$ are the sum of $$n$$ nonzero squares.

A corollary of Fermat's two-squares theorem is:

An integer greater than $$1$$ can be written as a sum of two squares if and only if its prime decomposition contains no prime congruent to $$3 \mod 4$$ raised to an odd power.

Legendre's three-square theorem states:

A natural number $$n$$ can be represented as the sum of three squares if and only if it is not of the form $$n = 4 a ( 8 b + 7 )$$ for integers $$a$$ and $$b$$.

• I am asking for general $n$ squares excluding the trivials. – user662075 Apr 8 at 14:12
• Don't forget Lagrange's four-square theorem: that every natural number can be represented as the sum of four integer squares. Though this does not appear to be what the OP is asking, mind you. – Mr Pie Apr 8 at 14:19