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Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ need four nonzero squares

Statement

Let $a \in N$ , let $m$ be a positive integer with $m \equiv 7 (m o d 8)$ (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ), and put $n = 4^{a} m$ , the power being the natural power in the commutative monoid $(Z, \cdot, 1)$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ). Then $n$ is a sum of four integer squares, and in every representation $(x_{1}, x_{2}, x_{3}, x_{4}) \in Z^{4}$ of $n$ (Representations as sums of four squares) all four coordinates are nonzero.

Facts & Assumptions

Given: A natural number $a$ , a positive integer $m$ with $m \equiv 7 (m o d 8)$ , and $n = 4^{a} m$ .

[F1]

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).

[L1]

Every nonnegative integer is a sum of four integer squares (Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares).

[L2]

For $a \in N$ and a positive integer $m$ with $m \equiv 7 (m o d 8)$ , there are no integers $x, y, z$ with $4^{a} m = x^{2} + y^{2} + z^{2}$ (Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares).

[L3]

In a monoid $(M, \cdot, e)$ the natural powers of $g \in M$ satisfy $g^{0} = e$ and $g^{σ (n)} = g^{n} \cdot g$ for $n \in N$ , where $σ$ is the successor on $N$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

$(Z, \cdot, 1)$ is a commutative monoid ( $(Z, \cdot, 1)$ is a commutative monoid whose group of units is ${1, - 1}$ ; equivalently $u ∣ 1$ holds exactly for $u = 1$ and $u = - 1$ ).

[L5]

The order on $Z$ is total, $x \leq y$ implies $x + z \leq y + z$ , and $0 < x$ , $0 < y$ imply $0 < x y$ (The integers form a totally ordered ring).

[L6]

The embedding of $N$ into $Z$ is injective, preserves order, and has image exactly the nonnegative integers (The naturals embed in the integers).

[L7]

On $N$ , the strict order is membership and $0 \in n$ whenever $n \neq 0$ (On $N$ the order is membership: $m < n ⟺ m \in n$ ).

[L8]

One has $m < n$ exactly when $σ (m) \leq n$ (Discreteness: $σ (n)$ is the immediate successor), and $1 = σ (0)$ (The natural numbers $N$ (von Neumann)).

[L9]

Let $S \subseteq N$ . If $0 \in S$ and $σ (n) \in S$ whenever $n \in S$ , then $S = N$ (The principle of mathematical induction).

Proof

technique · contradiction

1.1

The integers satisfy $0 < 1$ , and every positive integer is at least $1$ : the first because the embedded natural number $1$ is nonnegative and differs from the embedded natural number $0$ , which is the integer $0$ , by injectivity in [L6]; and if $y > 0$ then [L6] writes $y$ as the image of a unique natural $t$ , with $t \neq 0$ , so [L7] gives $0 \in t$ , hence $0 < t$ , and [L8] turns this into $1 = σ (0) \leq t$ , which order preservation in [L6] carries to $1 \leq y$ .

L6L7L8

1.2

Suppose, for contradiction, that some quadruple $(x_{1}, x_{2}, x_{3}, x_{4}) \in Z^{4}$ satisfies $n = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}$ with $x_{j} = 0$ for at least one index $j$ .

assume-contra

2.1

Let $U = {a \in N : 4^{a} \geq 1}$ . Since $4^{0} = 1$ by [L3] in the monoid of [L4], the set $U$ contains $0$ . If $a \in U$ , then $4^{σ (a)} = 4^{a} \cdot 4$ by [L3], and both factors are at least $1$ : $4^{a} \geq 1$ by the hypothesis $a \in U$ , and step 1.1 gives $4 \geq 1$ because $4$ is a positive integer. Thus both factors are positive, so [L5] gives $4^{σ (a)} > 0$ , and step 1.1 then gives $4^{σ (a)} \geq 1$ . Hence $σ (a) \in U$ , and [L9] yields $U = N$ .

step 1.1L3L4L5L9

3.1

Since $m$ is positive, step 1.1 gives $m \geq 1$ ; and step 2.1 gives $4^{a} \geq 1$ . So both factors in $n = 4^{a} m$ are positive, [L5] gives $n > 0$ , and in particular $n$ is nonnegative.

step 1.1step 2.1L5

4.1

By [L1] applied to the nonnegative integer $n$ of step 3.1, $n$ is a sum of four integer squares, so a representation in the sense of [F1] exists.

step 3.1L1F1

5.1

Under the assumption of step 1.2, deleting the coordinate $x_{j}$ leaves three integers $y_{1}, y_{2}, y_{3}$ , the other coordinates in any order, with $y_{1}^{2} + y_{2}^{2} + y_{3}^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} - x_{j}^{2} = n = 4^{a} m$ , which [L2] excludes; the assumption therefore fails, so every representation of $n$ has all four coordinates nonzero, and by step 4.1 at least one representation exists.

step 4.1step 1.2L2F1algebradischarge-contradiction∎

Remarks

What the two clauses say together. Four squares suffice for $n$ , by Lagrange's four-square theorem: every nonnegative integer is a sum of four integer squares, and three do not, by Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares; the second clause is the sharper form of the latter, since a representation with a zero coordinate is exactly a representation of $n$ by three squares with a fourth coordinate added. So for these $n$ the number four in Lagrange's theorem cannot be lowered.

The smallest instances. Taking $a = 0$ and $m = 7$ gives $n = 7$ ; taking $a = 1$ and $m = 7$ gives $n = 28$ ; and taking $a = 0$ and $m = 15$ gives $n = 15$ , so the statement is not about $7$ alone.

Depends on

Used by

7 and 28 have four-square representations with no zero coordinate Example

Dependency tree · two levels

60 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

Evan Dummit, Number Theory (part 9): The Geometry of Numbers, §9.1.3 (standard reference, not scraped)
Karl-Dieter Crisman, Number Theory: In Context and Interactive, §14.2, Fact 14.2.1 (standard reference, not scraped)

Positive integers 4am4am with m≡7(mod8)m≡7(mod8) need four nonzero squares