Prove there are infinitely many positive integers which cannot be represented as a sum of four non-zero squares. Prove there are infinitely many positive integers which cannot be represented as a sum of four non-zero squares. Every positive integer can be written as the sum of four squares. But not all necessarily non-zero. Any hints on this?

Assume there are $4$ such non-zero squares that add up to $2^{2n+1}$ for any $n \ge 1$, i.e., you have
$$2^{2n+1} = a^2 + b^2 + c^2 + d^2 \tag{1}\label{eq1A}$$
However, note all perfect squares are congruent to $0$, $1$ or $4$ modulo $8$. Since you just asked for a hint, the rest of the answer is in the spoiler below.

 Any positive integer of the form $2^{k}$ where $k \ge 3$, such as where $k = 2n + 1$ for $n \ge 1$, is congruent to $0$ modulo $8$ and can only be the sum of $4$ squares if they are all even (since all $4$ odd gives a congruence of $4$ modulo $8$, $3$ odd gives $3$ or $7$, $2$ odd gives $2$ or $6$, and just $1$ odd gives $1$ or $5$). Thus, you have $a = 2a_1$, $b = 2b_1$, $c = 2c_1$ and $d = 2d_1$. Substituting this into \eqref{eq1A} and dividing both sides by $4$ gives

 $$2^{2(n-1) + 1} = 2^{2n - 1} = a_1^2 + b_1^2 + c_1^2 + d_1^2 \tag{2}\label{eq2A}$$

 This is an equation of the same form, so as long as the power of $2$ is $\ge 3$, you can repeat the procedure. Repeating this $n$ times gives

 $$2^{2(n-n) + 1} = 2^{1} = a_n^2 + b_n^2 + c_n^2 + d_n^2 \tag{3}\label{eq3A}$$

 This is not possible since the RHS is at least $4$ but the LHS is just $2$. This means at least one (actually, $2$) of the squares in \eqref{eq1A} must have been $0$. Since \eqref{eq1A} only required that $n \ge 1$, and \eqref{eq3A} shows it works for $n = 0$ also, you have an infinite # of positive integers of the form $2^{2n+1}$ which cannot be represented as the sum of $4$ non-zero squares.
 
 Note you can also use induction to prove $2^{2n+1} \; \forall \; n \ge 0$ cannot be represented by a sum of $4$ non-zero squares by using \eqref{eq3A} as the base case, and then using the modulo $8$ congruences to show you can reduce the $n = k + 1$ case to the $n = k$ case in the inductive step.

