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The two intersecting square sets modulo $11$ give $1^{2} + 3^{2} + 1 = 11$

Example

Let $p = 11$ and work with the representatives $0, 1, \dots, 10$ . Put

$Q = {r : 0 \leq r \leq 10 and r \equiv y^{2} (m o d 11) for some y \in Z}, R = {r : 0 \leq r \leq 10 and r \equiv - 1 - y^{2} (m o d 11) for some y \in Z} .$

Then $Q = {0, 1, 3, 4, 5, 9}$ and $R = {1, 5, 6, 7, 9, 10}$ , each with $6 = (11 + 1) / 2$ elements, and $Q \cap R = {1, 5, 9}$ . Taking the common value $1$ , which is $1^{2}$ and is also congruent to $- 1 - 3^{2}$ , gives

$1^{2} + 3^{2} + 1 = 11 \equiv 0 (m o d 11) .$

This is the mechanism behind Every nonzero residue modulo an odd prime is a sum of two squares at $a = - 1$ , written out at $p = 11$ : two subsets of the $11$ residues, each of size $6$ , cannot be disjoint. That proposition supplies such a common value for every odd prime, and For every prime $p$ the congruence $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ is solvable cites it rather than repeating the count.

Facts & Assumptions

Given: The prime $p = 11$ and the sets $Q$ and $R$ displayed above.

[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 every prime $p$ there are integers $x, y$ with $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ (For every prime $p$ the congruence $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ is solvable).

[L2]

Let $p$ be an odd prime and let $a \in Z$ with $p ∤ a$ . Then there are integers $x, y$ such that $x^{2} + y^{2} \equiv a (m o d p)$ (Every nonzero residue modulo an odd prime is a sum of two squares).

Verification

technique · direct

1.1

Since $y$ and $- y$ have the same square and $y \equiv y + 11$ modulo $11$ , every square residue is $y^{2}$ for some $y$ with $0 \leq y \leq 5$ ; computing, $0^{2} = 0$ , $1^{2} = 1$ , $2^{2} = 4$ , $3^{2} = 9$ , $4^{2} = 16 \equiv 5$ and $5^{2} = 25 \equiv 3$ , so $Q = {0, 1, 3, 4, 5, 9}$ .

givenF1algebra

2.1

For the same six values of $y$ , $- 1 - 0 = - 1 \equiv 10$ , $- 1 - 1 = - 2 \equiv 9$ , $- 1 - 4 = - 5 \equiv 6$ , $- 1 - 9 = - 10 \equiv 1$ , $- 1 - 16 = - 17 \equiv 5$ and $- 1 - 25 = - 26 \equiv 7$ modulo $11$ , so $R = {1, 5, 6, 7, 9, 10}$ .

step 1.1F1algebra

3.1

Comparing the two displayed lists, $Q \cap R = {1, 5, 9}$ ; taking $r = 1$ with $1 \equiv 1^{2}$ from step 1.1 and $1 \equiv - 1 - 3^{2}$ from step 2.1 gives $1^{2} + 3^{2} \equiv - 1 (m o d 11)$ , hence $1^{2} + 3^{2} + 1 = 1 + 9 + 1 = 11 \equiv 0 (m o d 11)$ , an instance of [L1] at $p = 11$ .

step 1.1step 2.1L1F1algebra

4.1

The nonempty intersection exhibited in step 3.1 is exactly what [L2] asserts at $a = - 1$ for an arbitrary odd prime, since $R$ is the set of residues of $a - y^{2}$ at $a = - 1$ ; the argument at $p = 11$ therefore instantiates the cited proposition rather than adding to it.

step 3.1L2∎

Remarks

Why $11$ is a good modulus for this. Here $11 \equiv 3 (m o d 4)$ and $- 1 \equiv 10$ is not in $Q$ , so $- 1$ is not itself a square modulo $11$ and no pair with $y = 0$ works. The example is therefore not a disguised one-square case: two squares are genuinely needed, which is what the cited proposition provides.

Other solutions exist. The intersection has the further common values $5$ and $9$ : the value $5$ is $4^{2}$ and also $- 1 - 4^{2}$ modulo $11$ , giving $4^{2} + 4^{2} + 1 = 33 = 3 \cdot 11$ , and the value $9$ is $3^{2}$ and also $- 1 - 1^{2}$ , giving $3^{2} + 1^{2} + 1 = 11$ . So the solution pair is not unique; the cited proposition asserts that at least one common value exists and says nothing about how many do.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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, Lemma 6.4 (standard reference, not scraped)

The two intersecting square sets modulo 1111 give 12+32+1=1112+32+1=11