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Lagrange's four square_theorem states that every positive integer $N$ can be written as a sum of four squares of integers. At present, let's focus only on positive integer summands; that is, $N=a_1^2+a_2^2+a_3^2+a_4^2$ with $a_i\in\mathbb{N}$.

If now mix-up sums and products, something rather curious happens: the greatest common divisor (gcd) becomes low-primed. I wish to know why.

CLAIM. If $n\geq26$ is an integer, then $$\text{GCD}\,\left\{a_1a_2a_3a_4\,: \, 4n+1=a_1^2+a_2^2+a_3^2+a_4^2 \,\,\, \text{and $a_i\in\mathbb{Z}^{+}$}\right\}=2^b3^c$$ for some $b+c>0$.

For example, \begin{align} 4(26)+1&=7^2+6^2+4^2+2^2=8^2+4^2+4^2+3^2 \\ &=8^2+6^2+2^2+1^2=9^2+4^2+2^2+2^2. \end{align} Therefore, the GCD equals $2^43^1$.

UPDATED (later November 29, 2016).

I now have a more specific claim for the specific powers of $2$ and $3$, showing that even these numbers are low-powered.

CONJECTURE. Denote $a=a(n)$ and $b=b(n)$ the exponents $2^b3^c$ from above. Then, $$b(n)=\begin{cases} 4 \qquad \text{if $n$ is even} \\ 3 \qquad \text{if $n$ is odd}; \end{cases}$$ $$c(n)=\begin{cases} 0 \qquad \text{if $n=3k$} \\ 2 \qquad \text{if $n=3k+1$} \\ 1 \qquad \text{if $n=3k+2$}. \end{cases}$$

"It's easy for number theory to be hard." - anonymous.

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    $\begingroup$ In other words, given a prime $p>3$ we want to find four squares non-divisible by $p$ which sum up to $4n+1$. This is possible at least modulo $p$ by standard reasons like Cauchy-Davenport. We may try to count the number of representations of $n$ as $n=x^2+y^2+z^2+p^2t^2$ (how to do it?) and prove that it is strictly less than the number of representations as a sum of four squares. Alternatively, we may try to modify the inductive proof of Lagrange theorem carefully avoiding divisibility by $p$. $\endgroup$
    – Fedor Petrov
    Commented Nov 29, 2016 at 12:38
  • $\begingroup$ Uhmm... interesting. $\endgroup$
    – T. Amdeberhan
    Commented Nov 29, 2016 at 12:45
  • $\begingroup$ @T.Amdeberhan Do you know of any example for $n \geq 26$ where none of $a_1,a_2,a_3,a_4$ is divisible by 3? $\endgroup$
    – Farewell
    Commented Nov 29, 2016 at 13:23
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    $\begingroup$ Surely this disappears with large n? Take any m which has two representations as sums of two squares, and for M a sufficiently large multiple of 4 also a sum of two squares that M+m has two representations with large gcd? Or are you taking gcd over all decompositions? Gerhard "Is Squaring Up Both Decomposition" Paseman, 2016.11.29. $\endgroup$
    – Gerhard Paseman
    Commented Nov 29, 2016 at 16:20
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    $\begingroup$ GCD over all decompositions. $\endgroup$
    – T. Amdeberhan
    Commented Nov 29, 2016 at 16:40

1 Answer 1

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The claim is certainly true for $n$ sufficiently large, and "sufficiently large" could be specified explicitly with more care.

We follow the suggestion of Fedor Petrov, and rely on the results of Brüdern & Fouvry (J. reine angew. Math. 454 (1994), 59-96) and of Heath-Brown & Tolev (J. reine angew. Math. 558 (2003), 159-224). These yield, for any prime $p\geq 5$, that if we count the representations $$4n+1=a_1^2+a_2^2+a_3^2+a_4^2$$ smoothly (with positive integers $a_i$), then the proportion of representations such that $p\mid a_1$ is at most $1/(p-1)+o(1)$ as $n\to\infty$, where $o(1)$ is uniform in $p$ (note that we can restrict to $p\ll n^{1/2}$). See especially Theorem 3 and Lemmata 6-7 in Brüdern & Fouvry, and the displays (340), (346), (349), (352) in Heath-Brown & Tolev. It follows that for $p\geq 7$ the proportion of representations such that $p\mid a_1a_2a_3a_4$ is at most $2/3+o(1)$, hence for $n$ sufficiently large (independent of $p$) there is a representation such that $p\nmid a_1a_2a_3a_4$. For $p=5$ we need to be more careful, because $4/(p-1)$ equals $1$ in this case, and subtract the (positive) proportion of representations such that two of the $a_i$'s are divisible by $p$. This is also covered by the mentioned works as they discuss general divisibility constraints $d_i\mid a_i$.

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  • $\begingroup$ Does this work for numbers $4n+3$ instead of $4n+1$? $\endgroup$
    – Fedor Petrov
    Commented Nov 29, 2016 at 20:57
  • $\begingroup$ @FedorPetrov: I think this works for arbitrary numbers $N$ (instead of $N=4n+1$) as long as the odd part of $N$ exceeds $N^c$ for some fixed $c>3/4$. $\endgroup$
    – GH from MO
    Commented Nov 29, 2016 at 21:35
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    $\begingroup$ If it works for large enough odd numbers, it works for all numbers with large enough odd part by some obvious reasons (multiplying by 2 works as $2(a^2+b^2+c^2+d^2)=(a+b)^2+(a-b)^2+(c+d)^2+(c-d)^2$, and the greatest odd common divisor of $a+b,a-b,c+d,c-d$ is the same as for $a,b,c,d$). $\endgroup$
    – Fedor Petrov
    Commented Nov 29, 2016 at 22:31
  • $\begingroup$ @FedorPetrov: I think you are right. I focused on the actual proportions in my post, and it is for the $o(1)$ terms that we need a sufficiently large odd part of $N$. But in the end, as you say, the $2$-part of $N$ is irrelevant for the odd gcd of the products $a_1a_2a_3a_4$. $\endgroup$
    – GH from MO
    Commented Nov 29, 2016 at 23:23
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    $\begingroup$ @T.Amdeberhan: Specifying an explicit lower bound for $n$ seems doable, but it will involve considerable amount of work. The main task is to calculate explicitly the implied constants in (349) and (352) of Heath-Brown & Tolev, and for that (I think) one needs to study a big portion of the paper in detail. If you are that much interested in this problem, use it as motivation to study Heath-Brown & Tolev. $\endgroup$
    – GH from MO
    Commented Nov 30, 2016 at 16:13

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