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Some multiple $p m$ with $1 \leq m < p$ is a sum of four squares

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ). Then there is an integer $m$ with $1 \leq m < p$ for which $p m$ is a sum of four integer squares (Representations as sums of four squares).

Facts & Assumptions

Given: A prime $p$ .

[F1]

An integer $p$ is prime when $p > 1$ and $d ∣ p$ with $d > 0$ force $d = 1$ or $d = p$ ; in words, $p$ exceeds $1$ , and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[F2]

A representation of a nonnegative integer $n$ as a sum of four squares is an ordered quadruple $(a, b, c, d) \in Z^{4}$ with $n = a^{2} + b^{2} + c^{2} + d^{2}$ (Representations as sums of four squares).

[F3]

For $a, b, n \in Z$ , $a \equiv b (m o d n)$ means $n ∣ (a - b)$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[F4]

For $d, a \in Z$ , $d ∣ a$ means $a = d q$ for some $q \in Z$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L1]

For every prime $p$ there are integers $x, y$ with $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ (For every prime $p$ the congruence $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ is solvable).

[L2]

For an integer $m \geq 1$ and $a \in Z$ there is exactly one integer $r$ with $a \equiv r (m o d m)$ and $- m < 2 r \leq m$ , and consequently $4 r^{2} \leq m^{2}$ (The least absolute remainder modulo a positive integer).

[L3]

If $a \equiv a^{'} (m o d n)$ and $b \equiv b^{'} (m o d n)$ , then $a + b \equiv a^{'} + b^{'} (m o d n)$ and $a b \equiv a^{'} b^{'} (m o d n)$ (Congruent integers may be added, subtracted and multiplied: representative changes preserve both arithmetic operations).

Proof

technique · direct

1.1

Fix integers $u, v$ with $u^{2} + v^{2} + 1 \equiv 0 (m o d p)$ , which [L1] supplies for the prime $p$ .

givenL1choose

2.1

Since $p > 1$ by [F1], the modulus $p$ satisfies the hypothesis of [L2]; let $x$ and $y$ be the least absolute remainders of $u$ and $v$ modulo $p$ , so that $x \equiv u (m o d p)$ , $y \equiv v (m o d p)$ , $4 x^{2} \leq p^{2}$ and $4 y^{2} \leq p^{2}$ .

step 1.1F1L2construct

3.1

By [L3] applied to the products $x \cdot x$ and $y \cdot y$ and then to the sums, $x^{2} + y^{2} + 1 \equiv u^{2} + v^{2} + 1 (m o d p)$ , and $u^{2} + v^{2} + 1 \equiv 0 (m o d p)$ ; hence $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ , which by [F3] and [F4] says $p ∣ x^{2} + y^{2} + 1$ .

step 1.1step 2.1L3F3F4algebra

4.1

Write $x^{2} + y^{2} + 1 = p m$ with $m \in Z$ , as [F4] permits; the left-hand side is at least $1$ because $x^{2} \geq 0$ and $y^{2} \geq 0$ , and $p > 0$ , so $m \geq 1$ .

step 3.1F4algebra

5.1

Multiplying step 4.1 by $4$ and using the two bounds of step 2.1 gives $4 p m = 4 x^{2} + 4 y^{2} + 4 \leq p^{2} + p^{2} + 4 = 2 p^{2} + 4$ ; since $p \geq 2$ by [F1] we have $p^{2} \geq 4$ , so $2 p^{2} + 4 \leq 3 p^{2} < 4 p^{2}$ , whence $4 p m < 4 p^{2}$ , $p m < p^{2}$ and $m < p$ .

step 2.1step 4.1F1algebra

6.1

The equation $p m = x^{2} + y^{2} + 1^{2} + 0^{2}$ of step 4.1 exhibits the quadruple $(x, y, 1, 0) \in Z^{4}$ as a representation of $p m$ , so $p m$ is a sum of four integer squares with $1 \leq m < p$ .

step 4.1step 5.1F2∎

Remarks

Why the centring is needed. The pair $(u, v)$ produced by [L1] is subject to no size condition, so $u^{2} + v^{2} + 1$ can be an arbitrarily large multiple of $p$ . Replacing $u, v$ by their least absolute remainders leaves the congruence class untouched and buys the two bounds $4 x^{2} \leq p^{2}$ and $4 y^{2} \leq p^{2}$ , which is what forces the multiplier below $p$ .

The bound is not tight, and does not need to be. What the coordinate bounds give at step 5.1 is $4 p m \leq 2 p^{2} + 4$ , and only the weaker $4 p m < 4 p^{2}$ is used. The comparison $2 p^{2} + 4 < 4 p^{2}$ needs just $p^{2} > 2$ , which every prime satisfies, so $p = 2$ needs no separate treatment even though the coordinate bound $4 x^{2} \leq p^{2}$ can be attained there, at $x = \pm 1$ .

The case $m = 1$ . Nothing excludes it, and it is the case $p = x^{2} + y^{2} + 1$ in which the prime is already a sum of four squares.

Depends on

Used by

Every prime is a sum of four integer squares Theorem

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Keith Conrad, Proofs by Descent, §6, Theorem 6.6 (Step 1) (standard reference, not scraped)
MIT 18.781 Theory of Numbers, Lecture 22, Lemma 81 (standard reference, not scraped)

Some multiple pmpm with 1≤m<p1≤m<p is a sum of four squares