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Descent step: a smaller multiple of $p$ 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 $p$ ), let $m$ be an integer with $1 < m < p$ , and suppose $p m$ is a sum of four integer squares (Representations as sums of four squares). Then there is an integer $n$ with $1 \leq n < m$ for which $p n$ is a sum of four integer squares.

Facts & Assumptions

Given: A prime $p$ , an integer $m$ with $1 < m < p$ , and a representation of $p m$ as a sum of four integer squares.

[F1]

An integer $p$ is prime when $p > 1$ and $d ∣ p$ with $d > 0$ force $d = 1$ or $d = p$ (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 all integers $x_{1}, \dots, x_{4}, y_{1}, \dots, y_{4}$ , setting $z_{1} = x_{1} y_{1} + x_{2} y_{2} + x_{3} y_{3} + x_{4} y_{4}$ , $z_{2} = x_{1} y_{2} - x_{2} y_{1} - x_{3} y_{4} + x_{4} y_{3}$ , $z_{3} = x_{1} y_{3} + x_{2} y_{4} - x_{3} y_{1} - x_{4} y_{2}$ and $z_{4} = x_{1} y_{4} - x_{2} y_{3} + x_{3} y_{2} - x_{4} y_{1}$ gives $(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}) (y_{1}^{2} + y_{2}^{2} + y_{3}^{2} + y_{4}^{2}) = z_{1}^{2} + z_{2}^{2} + z_{3}^{2} + z_{4}^{2}$ (Euler's four-square product identity).

[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 $p$ is prime, $1 < m < p$ and $p m = a^{2} + b^{2} + c^{2} + d^{2}$ for integers $a, b, c, d$ , then the least absolute remainders $a^{'}, b^{'}, c^{'}, d^{'}$ of $a, b, c, d$ modulo $m$ satisfy $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} = m n$ for an integer $n$ with $1 \leq n < m$ (The centred residue quadruple of $p m = a^{2} + b^{2} + c^{2} + d^{2}$ has norm $m n$ with $1 \leq n < m$ ).

[L4]

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

[L5]

If $x, y \in Z$ are nonzero then $x y \neq 0$ ; consequently, if $x z = y z$ and $z \neq 0$ , then $x = y$ (The integers have no zero divisors; multiplicative cancellation).

Proof

technique · direct

1.1givenF2choose

Fix integers $a, b, c, d$ with $p m = a^{2} + b^{2} + c^{2} + d^{2}$ , which the hypothesis supplies.

2.1step 1.1L2construct

Since $m > 1$ , [L2] applies with modulus $m$ ; let $a^{'}, b^{'}, c^{'}, d^{'}$ be the least absolute remainders of $a, b, c, d$ modulo $m$ , so $a \equiv a^{'}$ , $b \equiv b^{'}$ , $c \equiv c^{'}$ and $d \equiv d^{'}$ modulo $m$ .

3.1step 1.1step 2.1L3F1

By [L3] applied to $p$ , $m$ and the representation of step 1.1, there is an integer $n$ with $1 \leq n < m$ and $a^{' 2} + b^{' 2} + c^{' 2} + d^{' 2} = m n$ .

3.2step 1.1step 2.1L4F3algebra

Put $A = a a^{'} + b b^{'} + c c^{'} + d d^{'}$ ; substituting the congruences of step 2.1 and using [L4] gives $A \equiv a^{2} + b^{2} + c^{2} + d^{2} (m o d m)$ , and $a^{2} + b^{2} + c^{2} + d^{2} = p m \equiv 0 (m o d m)$ by [F3], so $A \equiv 0 (m o d m)$ .

3.3step 2.1L4F3algebra

Put $B = a b^{'} - b a^{'} - c d^{'} + d c^{'}$ , $C = a c^{'} + b d^{'} - c a^{'} - d b^{'}$ and $D = a d^{'} - b c^{'} + c b^{'} - d a^{'}$ ; the same substitution and [L4] give $B \equiv a b - b a - c d + d c = 0$ , $C \equiv a c + b d - c a - d b = 0$ and $D \equiv a d - b c + c b - d a = 0$ modulo $m$ .

4.1step 1.1step 2.1step 3.1L1

Applying [L1] to $x = (a, b, c, d)$ and $y = (a^{'}, b^{'}, c^{'}, d^{'})$ produces exactly the four quantities $A, B, C, D$ of steps 3.2 and 3.3 as $z_{1}, z_{2}, z_{3}, z_{4}$ , so $(p m) (m n) = A^{2} + B^{2} + C^{2} + D^{2}$ .

4.2step 3.2step 3.3F3F4construct

By [F3] and [F4] the congruences of steps 3.2 and 3.3 say $m ∣ A$ , $m ∣ B$ , $m ∣ C$ and $m ∣ D$ ; write $A = m A_{1}$ , $B = m B_{1}$ , $C = m C_{1}$ , $D = m D_{1}$ with $A_{1}, B_{1}, C_{1}, D_{1} \in Z$ .

5.1step 4.1step 4.2algebra

Substituting step 4.2 into step 4.1 gives $p m^{2} n = m^{2} (A_{1}^{2} + B_{1}^{2} + C_{1}^{2} + D_{1}^{2})$ .

6.1step 5.1L5algebra

Since $m > 1$ is nonzero, $m^{2} \neq 0$ by [L5], so cancelling $m^{2}$ in step 5.1 by [L5] gives $p n = A_{1}^{2} + B_{1}^{2} + C_{1}^{2} + D_{1}^{2}$ .

7.1step 3.1step 6.1F2∎

The quadruple $(A_{1}, B_{1}, C_{1}, D_{1}) \in Z^{4}$ therefore represents $p n$ as a sum of four integer squares, with $1 \leq n < m$ from step 3.1.

Remarks

The sign pattern is doing the work. Steps 3.2 and 3.3 substitute $a^{'} \equiv a$ , $b^{'} \equiv b$ , $c^{'} \equiv c$ , $d^{'} \equiv d$ into the four bilinear forms of [L1] and read off that all four become divisible by $m$ : the first because it becomes the norm $p m$ , the other three because they become expressions in which the terms cancel identically. That is a property of this particular choice of signs, and Why the descent fixes one sign pattern in the four-square identity records which other choices share it.

The hypothesis $m > 1$ is used twice. It gives the modulus of [L2] and it is part of what [L3] needs; and it is not a restriction in practice, since $m = 1$ is the case in which $p$ itself is already a sum of four squares and no descent is wanted.

Depends on

Used by

Dependency tree · two levels

26 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 2) (standard reference, not scraped)
MIT 18.781 Theory of Numbers, Lecture 22, Theorem 80 (standard reference, not scraped)

Descent step: a smaller multiple of pp is a sum of four squares