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Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares

Statement

Let $a \in N$ and 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 $1$ ). Then there are no integers $x, y, z$ with $4^{a} m = x^{2} + y^{2} + z^{2}$ , where $4^{a}$ is the natural power of $4$ 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 $u = - 1$ ).

Facts & Assumptions

Given: A positive integer $m$ with $m \equiv 7 (m o d 8)$ .

[F1]

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 $1$ ).

[F2]

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]

There are no integers $x, y, z$ with $x^{2} + y^{2} + z^{2} \equiv 7 (m o d 8)$ (No sum of three integer squares is congruent to $7$ modulo $8$ ).

[L2]

If $x, y, z \in Z$ and $4 ∣ x^{2} + y^{2} + z^{2}$ , then $x$ , $y$ and $z$ are all even (If $4$ divides $x^{2} + y^{2} + z^{2}$ then $x$ , $y$ and $z$ are all even).

[L3]

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).

[L4]

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 $g^{0} = e$ ).

[L5]

$(Z, \cdot, 1)$ 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 $u = - 1$ ).

[L6]

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 · induction

1.1

Let $S$ be the set of $a \in N$ such that for every positive integer $m$ with $m \equiv 7 (m o d 8)$ there are no integers $x, y, z$ with $4^{a} m = x^{2} + y^{2} + z^{2}$ .

givenconstruct

1.2

Base case $0 \in S$ : by [L4] in the monoid of [L5], $4^{0} = 1$ , so $4^{0} m = m$ and a representation $m = x^{2} + y^{2} + z^{2}$ would give $x^{2} + y^{2} + z^{2} \equiv 7 (m o d 8)$ by the hypothesis $m \equiv 7 (m o d 8)$ , which [L1] excludes.

baseL1L4L5F1algebra

2.1

Induction step: let $a \in S$ , let $m$ be a positive integer with $m \equiv 7 (m o d 8)$ , and suppose integers $x, y, z$ satisfy $4^{σ (a)} m = x^{2} + y^{2} + z^{2}$ ; by [L4] and [L5], $4^{σ (a)} = 4^{a} \cdot 4$ , so $x^{2} + y^{2} + z^{2} = 4 \cdot (4^{a} m)$ and hence $4 ∣ x^{2} + y^{2} + z^{2}$ by [F2].

step 1.1ihL4L5F2algebra

3.1

By [L2] the coordinates $x$ , $y$ , $z$ are then all even, so $x = 2 x^{'}$ , $y = 2 y^{'}$ and $z = 2 z^{'}$ for integers $x^{'}, y^{'}, z^{'}$ .

step 2.1L2construct

4.1

Substituting gives $4 \cdot (4^{a} m) = x^{2} + y^{2} + z^{2} = 4 (x^{' 2} + y^{' 2} + z^{' 2})$ , and cancelling the nonzero factor $4$ by [L3] yields $4^{a} m = x^{' 2} + y^{' 2} + z^{' 2}$ , which contradicts $a \in S$ since $m$ is a positive integer congruent to $7$ modulo $8$ .

step 2.1step 3.1L3algebra

5.1

So no such $x, y, z$ exist and $σ (a) \in S$ ; with the base case of step 1.2, [L6] gives $S = N$ , which is the assertion.

step 1.2step 4.1L6discharge-induction∎

Remarks

Three descriptions of the same integers. For a positive $m$ , the condition $m \equiv 7 (m o d 8)$ says $m = 8 b + 7$ for an integer $b$ , and $b \geq 0$ because $8 b + 7 > 0$ ; so the integers excluded here are exactly those of the form $4^{a} (8 b + 7)$ with $a$ a natural number and $b$ a nonnegative integer, which is how Dummit writes them. Crisman's phrase for the same set, an even power of two times an odd number congruent to seven modulo eight, is a third description: $4^{a} = 2^{2 a}$ and every $m \equiv 7 (m o d 8)$ is odd.

Only one direction is proved. The statement says these integers are not sums of three squares. Its converse, that every other nonnegative integer is a sum of three squares, is Legendre's three-square theorem; it is not available from this page's declared prerequisites, and nothing here uses it. In particular the argument above rules out no integer beyond the ones named.

Why the induction is on the exponent. The base case is a congruence computation modulo $8$ and nothing more. The step is where the work is: it needs that a sum of three squares divisible by $4$ has all coordinates even, which is If $4$ divides $x^{2} + y^{2} + z^{2}$ then $x$ , $y$ and $z$ are all even, since without it the halved coordinates need not be integers.

Depends on

Used by

Positive integers 4ᵃ m with m≡ 7 (mod 8) need four nonzero squares Corollary
3 and 5 are sums of three squares and 15 is not Counterexample
7 and 28 are not sums of three integer squares Counterexample
7 and 28 have four-square representations with no zero coordinate Example

Dependency tree · two levels

34 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) are not sums of three integer squares