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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?

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    $\begingroup$ All squares are nonnegative. The problem comes because you rule out $0$, so $1=1^2+0^2+0^2+0^2$ is not allowed. All positive integers can be written as the sum of four squares if you allow $0$, so the problem is to show that infinitely many cannot be written without $0$. $\endgroup$ – Ross Millikan Jun 10 '20 at 4:19
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    $\begingroup$ @RossMillikan yep, and i wonder how to prove there're infinitely many cannot be written as a sum of four non-zero squares? $\endgroup$ – Emma Johnson Jun 10 '20 at 4:22
  • $\begingroup$ @EmmaJohnson FYI, the Uniqueness section of Wikipedia's "Lagrange's four-square theorem" gives the complete list of positive integers which cannot be represented as a sum of four non-zero squares. $\endgroup$ – John Omielan Aug 27 '20 at 20:23
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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.

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    $\begingroup$ It may be of interest that the positive integers that aren't representable as the sum of four nonzero squares are $1,3,5,9,11,17,29,41$ and the integers of the form $4^a\cdot 2$, $4^a\cdot 6$, $4^a\cdot 14$. And all positive integers except a few (twelve or fifteen, the largest of which is $33$, if I'm not misremembering) small ones are representable as the sum of five nonzero squares. $\endgroup$ – Daniel Fischer Jun 10 '20 at 14:48
  • $\begingroup$ +1. You could say it differently by supposing $n$ is the least $m\ge1$ such that $2^{2m+1}$ is an exception. Then $2^{2(n-1)+1}$ is not an exception, but $2^{2(n-1)+1}=a_1^2+b_1^2+c_1^2+d_1^2$ so $n-1\le 0.$ $\endgroup$ – DanielWainfleet Jun 10 '20 at 16:02
  • $\begingroup$ Couldn't this be generalized so that, if $k$ is any integer such that $2k$ is not representable as a sum of positive squares, then all numbers of the form $2^{2n+1}k$ also lack such a representation? $\endgroup$ – Cardioid_Ass_22 Jun 10 '20 at 19:00
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    $\begingroup$ @Cardioid_Ass_22 Yes, you're right. As Daniel Fischer's comment states with "$4^a\cdot 2, 4^a\cdot 6, 4^a\cdot 14$", it's true using your variables for $k = 1, 3, 7$. I haven't checked, but I assume that's probably all of them. $\endgroup$ – John Omielan Jun 10 '20 at 20:34

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