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

Example

The integers $7$ and $28$ have the four-square representations

$7 = 2^{2} + 1^{2} + 1^{2} + 1^{2}, 28 = 5^{2} + 1^{2} + 1^{2} + 1^{2},$

and in each of them every coordinate is nonzero. That is not an accident of these two displays: no representation of either integer can have a vanishing coordinate, because deleting a zero coordinate would exhibit the integer as a sum of three squares, which Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ are not sums of three integer squares forbids for $7 = 4^{0} \cdot 7$ and for $28 = 4^{1} \cdot 7$ . The two are the cases $a = 0$ and $a = 1$ of Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ need four nonzero squares at $m = 7$ .

Facts & Assumptions

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

[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).

[F2]

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 $m \equiv 7 (m o d 8)$ , in every representation $(x_{1}, x_{2}, x_{3}, x_{4}) \in Z^{4}$ of $4^{a} m$ as a sum of four integer squares all four coordinates are nonzero (Positive integers $4^{a} m$ with $m \equiv 7 (m o d 8)$ need four nonzero squares).

[L2]

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).

[L3]

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).

Verification

technique · direct

1.1

Both displays are representations in the sense of [F1], whose existence [L2] guarantees in advance: $4 + 1 + 1 + 1 = 7$ and $25 + 1 + 1 + 1 = 28$ ; and in each the four coordinates $2, 1, 1, 1$ and $5, 1, 1, 1$ are all nonzero.

givenF1L2algebra

1.2

The integer $7$ is positive and $8 ∣ 7 - 7$ , so $7 \equiv 7 (m o d 8)$ by [F2]; moreover $4^{0} \cdot 7 = 7$ and $4^{1} \cdot 7 = 28$ , so both integers have the form required by [L1] and [L3] with $m = 7$ and $a = 0$ or $a = 1$ .

givenF2algebra

2.1

If a representation of $7$ or of $28$ had a vanishing coordinate, deleting it would leave three integers whose squares sum to $7$ or to $28$ respectively, and by step 1.2 this is what [L3] excludes; so no representation of either integer has a vanishing coordinate.

step 1.1step 1.2L3algebra

3.1

Hence $7$ and $28$ each have a four-square representation, displayed in step 1.1, and every such representation has all four coordinates nonzero, which is [L1] at $a = 0$ and at $a = 1$ .

step 1.1step 1.2step 2.1L1F1∎

Remarks

Why both $a = 0$ and $a = 1$ are shown. The obstruction is stated for $4^{a} m$ , and its induction has a base case and a step. The witness $7$ exercises the base case and $28$ the first instance of the step, where the coordinates of a putative three-square representation are halved.

Depends on

Used by

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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 have four-square representations with no zero coordinate