LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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For every prime $p$ the congruence $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ is solvable

Statement

Let $p$ be a prime (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ). Then there are integers $x, y$ with $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ , the congruence being that of Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ .

Facts & Assumptions

Given: A prime $p$ .

[F1]

An integer $p$ is prime when $p > 1$ and $d ∣ p$ with $d > 0$ force $d = 1$ or $d = p$ ; in words, $p$ exceeds $1$ , and its only positive divisors are $1$ and $p$ (Prime and composite integers: $p$ is prime when $p > 1$ and its only positive divisors are $1$ and $p$ ).

[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 $1$ ).

[F3]

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]

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).

[L2]

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

[L3]

If $a \equiv a^{'} (m o d n)$ and $b \equiv b^{'} (m o d n)$ , then $a + b \equiv a^{'} + b^{'} (m o d n)$ , $a - b \equiv a^{'} - b^{'} (m o d n)$ and $a b \equiv a^{'} b^{'} (m o d n)$ (Congruent integers may be added, subtracted and multiplied: representative changes preserve both arithmetic operations).

Proof

technique · cases

1.1

If $2 ∣ p$ then $2$ is a positive divisor of $p$ , so [F1] forces $2 = 1$ or $2 = p$ , and $2 \neq 1$ leaves $p = 2$ ; hence either $p = 2$ or $p$ is odd, and these two cases exhaust the primes.

givenF1algebra

1.2

In the case $p = 2$ , take $x = 1$ and $y = 0$ : then $x^{2} + y^{2} + 1 = 1 + 0 + 1 = 2$ and $2 ∣ (2 - 0)$ , so $x^{2} + y^{2} + 1 \equiv 0 (m o d 2)$ .

assume-case twoF2algebra

1.3

In the case $p$ odd, $p ∤ - 1$ : otherwise $- 1 = p q$ for some integer $q$ by [F3], hence $1 = p (- q)$ , so $p ∣ 1$ and [L2] gives $p = 1$ or $p = - 1$ , both contradicting $p > 1$ from [F1].

assume-case oddF1F3L2algebra

2.1

For the odd case, apply [L1] to the odd prime $p$ with $a = - 1$ , whose hypothesis $p ∤ - 1$ is step 1.3: there are integers $x, y$ with $x^{2} + y^{2} \equiv - 1 (m o d p)$ .

step 1.3L1

3.1

For the odd case, $1 \equiv 1 (m o d p)$ since $p ∣ 0$ , so adding this to step 2.1 through [L3] gives $x^{2} + y^{2} + 1 \equiv - 1 + 1 = 0 (m o d p)$ .

step 2.1L3F2algebra

4.1

Both cases produce integers $x, y$ with $x^{2} + y^{2} + 1 \equiv 0 (m o d p)$ , and by step 1.1 no prime falls outside them.

step 1.2step 3.1cases-exhaustive∎

Remarks

Where the odd case comes from. For odd $p$ the work is done by Every nonzero residue modulo an odd prime is a sum of two squares at $a = - 1$ : the set $Q$ of square classes modulo $p$ , the zero class included, and the set of classes $- 1 - z$ for $z \in Q$ are two subsets of $Z / p$ with $(p + 1) / 2$ elements each, so they meet, and a common value gives $x^{2} \equiv - 1 - y^{2}$ . The hypothesis $p ∤ a$ of that proposition is what step 1.3 discharges, by an argument that does not use oddness; oddness is needed only to make the proposition applicable at all.

Why $p = 2$ is separate. The cited proposition is stated for odd primes, so it says nothing at $p = 2$ ; the pair $(1, 0)$ settles that case by computation rather than by weakening the proposition's hypothesis.

Depends on

Used by

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

Keith Conrad, Proofs by Descent, §6, Lemma 6.4 (standard reference, not scraped)
Evan Dummit, Number Theory (part 9): The Geometry of Numbers, §9.1.2, Lemma 2 (standard reference, not scraped)

For every prime pp the congruence x2+y2+1≡0(modp)x2+y2+1≡0(modp) is solvable