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LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

No sum of three integer squares is congruent to 7 modulo 8

Statement

There are no integers x,y,z with x2+y2+z2≡7(mod8) (Congruence modulo an integer: a≡b(modn) when n∣(a−b), including the moduli 0 and 1). In fact, the residues modulo 8 attained by sums of three integer squares are exactly 0,1,2,3,4,5,6.

Facts & Assumptions

Given: Integers x,y,z.

[F1]

For a,b,n∈Z, a≡b(modn) means n∣(a−b) (Congruence modulo an integer: a≡b(modn) when n∣(a−b), including the moduli 0 and 1).

[L1]

For every integer x: if x is odd then x2≡1(mod8), and if x is even then x2≡0(mod8) or x2≡4(mod8) (A square is congruent to 0, 1 or 4 modulo 8).

[L2]

If a≡a′(modn) and b≡b′(modn), then a+b≡a′+b′(modn) (Congruent integers may be added, subtracted and multiplied: representative changes preserve both arithmetic operations).

[L3]

For a,b∈Z with b≠0 there is exactly one pair (q,r) of integers with a=qb+r and 0≤r<∣b∣; moreover b∣a holds exactly when r=0 (Division with remainder for any nonzero divisor: for a∈Z and b≠0 there are unique q,r∈Z with a=qb+r and 0≤r<∣b∣).

Proof

technique · cases
1.1givenL3construct

By [L3] with b=2 each of x,y,z is even or odd; let t be how many of the three are odd, so t is 0, 1, 2 or 3, and these four values exhaust the possibilities.

1.2L1

By [L1], each odd coordinate contributes a square congruent to 1 modulo 8, and each even coordinate contributes a square congruent to 0 or to 4 modulo 8.

2.1step 1.1step 1.2assume-case threeL2F1algebra

If t=3 then, adding the three contributions by [L2], x2+y2+z2≡1+1+1=3(mod8).

2.2step 1.1step 1.2assume-case twoL2F1algebra

If t=2 then x2+y2+z2≡1+1+e(mod8) with e equal to 0 or 4, so the sum is congruent to 2 or to 6.

2.3step 1.1step 1.2assume-case oneL2F1algebra

If t=1 then x2+y2+z2≡1+e1+e2(mod8) with each ei equal to 0 or 4, giving 1, 5, 5 or 9; since 9≡1(mod8), the sum is congruent to 1 or to 5.

2.4step 1.1step 1.2assume-case zeroL2F1algebra

If t=0 then x2+y2+z2≡e1+e2+e3(mod8) with each ei equal to 0 or 4, giving 0, 4, 8 or 12; since 8≡0 and 12≡4 modulo 8, the sum is congruent to 0 or to 4.

3.1step 2.1step 2.2step 2.3step 2.4F1algebracases-exhaustive∎

Steps 2.1 to 2.4 cover the four values of t listed in step 1.1 and show that every sum of three squares is congruent to one of 0,1,2,3,4,5,6, never to 7, modulo 8. Conversely, the triples (0,0,0), (1,0,0), (1,1,0), (1,1,1), (2,0,0), (2,1,0) and (2,1,1) have sums of squares 0,1,2,3,4,5,6, respectively. Thus the attained residues are exactly the seven listed classes, and 7 is not attained.

Remarks

Why the cases are counted by parity rather than listed by value. Enumerating the possible triples of residues from {0,1,4} would give ten unordered choices; grouping them by how many coordinates are odd gives four, because the odd coordinates contribute a fixed residue and only the even ones branch. The exhaustiveness is then visible from step 1.1 alone.

Every listed residue is attained. Taking (x,y,z) to be (0,0,0), (1,0,0), (1,1,0), (1,1,1), (2,0,0), (2,1,0) and (2,1,1) gives sums 0,1,2,3,4,5,6, so the second sentence of the Statement is an equality of sets and not merely an inclusion.

Depends on

Used by

Dependency tree · two levels

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Sources