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$7$ and $28$ are not sums of three integer squares

Statement refuted

False claim: every nonnegative integer is a sum of three integer squares, that is, for every $n \geq 0$ there are integers $x, y, z$ with $n = x^{2} + y^{2} + z^{2}$ .

The integers $7$ and $28$ refute it. Both are excluded by Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares, as $7 = 4^{0} \cdot 7$ and $28 = 4^{1} \cdot 7$ with $7 \equiv 7 (m o d 8)$ ; and for $7$ the exclusion can also be seen by a finite search, which is carried out below so that the witness does not rest on the obstruction alone.

Facts & Assumptions

Given: The integers $7$ and $28$ .

[A1]

The false claim: for every integer $n \geq 0$ there are integers $x, y, z$ with $n = x^{2} + y^{2} + z^{2}$ .

[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 $1$ ).

[L1]

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).

[L2]

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 $8$ ).

[L3]

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).

Counterexample

technique · direct

1.1givenL2F1algebra

Since $8 ∣ 7 - 7$ , we have $7 \equiv 7 (m o d 8)$ by [F1], so a representation $7 = x^{2} + y^{2} + z^{2}$ would give $x^{2} + y^{2} + z^{2} \equiv 7 (m o d 8)$ , which [L2] excludes; hence $7$ is not a sum of three integer squares.

1.2givenalgebra

The same conclusion by finite search: in $7 = x^{2} + y^{2} + z^{2}$ each square is at most $7$ , so each lies in ${0, 1, 4}$ , and the sums of three members of that set are $0, 1, 2, 3, 4, 5, 6, 8, 9$ and $12$ , none of which is $7$ .

2.1step 1.1L3F1algebra

For $28$ : since $28 = 4 \cdot 7$ we have $4 ∣ 28$ , so a representation $28 = x^{2} + y^{2} + z^{2}$ would by [L3] have $x$ , $y$ and $z$ all even, say $x = 2 x^{'}$ , $y = 2 y^{'}$ , $z = 2 z^{'}$ ; then $4 \cdot 7 = 4 (x^{' 2} + y^{' 2} + z^{' 2})$ gives $7 = x^{' 2} + y^{' 2} + z^{' 2}$ , contradicting step 1.1.

3.1step 1.1step 1.2step 2.1A1L1∎

So neither $7$ nor $28$ is a sum of three integer squares, and [A1] is false; the two are the instances $a = 0$ and $a = 1$ of [L1] at $m = 7$ .

Remarks

Two independent routes for $7$ . Step 1.1 argues by residues modulo $8$ and step 1.2 by exhausting the finitely many candidates. The second uses no lemma about squares at all, so the non-existence claim does not depend on the congruence argument being right.

Why $28$ needs the halving step. The residue of $28$ modulo $8$ is $4$ , which is attainable by a sum of three squares, so the congruence argument does not apply to $28$ directly. It is the divisibility by $4$ , and the fact that this forces all three coordinates even, that reduces $28$ to $7$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Evan Dummit, Number Theory (part 9): The Geometry of Numbers, §9.1.3 (standard reference, not scraped)

77 and 2828 are not sums of three integer squares