Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
How statement and proof provenance work

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.

The normalized two-square count is multiplicative with the expected prime-power values

Statement

The arithmetic function f(n):=r2(n)/4 is multiplicative. More precisely,

f(2a)=1,

f(pa)=a+1for p≡1(mod4),

and

f(qa)={1,a even,0,a odd,for q≡3(mod4).

Facts & Assumptions

Given: A natural exponent a, a prime p≡1(mod4), a prime q≡3(mod4), and coprime positive integers m,n.

Proof

technique · direct
1.1givenalgebra

The ordered-sign representations of 1 are (±1,0) and (0,±1), so r2(1)=4 and therefore f(1)=1.

2.1step 1.1given

Write a positive integer N as N=2α∏i=1rpiei∏j=1sqjbj, where the pi≡1(mod4), the qj≡3(mod4), and all listed primes are distinct. If some bj is odd and N=x2+y2, write N=qj2k+1M with qj∤M. Repeatedly applying A prime congruent to 3 modulo 4 divides both coordinates of a divisible two-square sum shows that x=qjkx0 and y=qjky0, so qjM=x02+y02. One more application of the same lemma forces qj∣x0 and qj∣y0, hence qj2∣qjM, contradiction. Therefore r2(N)=0 whenever some bj is odd. If every bj is even, the cited Hackman theorem gives r2(N)4=∏i=1r(ei+1). Applying this to N=2a, N=pa, and N=qa yields r2(2a)4=1,r2(pa)4=a+1,r2(qa)4={1,a even,0,a odd.

3.1step 1.1step 2.1givenalgebra

Let f(n):=r2(n)/4. If gcd⁡(m,n)=1, the prime supports of m and n are disjoint. If step 2.1 gives f(m)=0 or f(n)=0, then some prime q≡3(mod4) has odd exponent in one factor, hence still odd exponent in mn, so step 2.1 also gives f(mn)=0=f(m)f(n). Otherwise every prime q≡3(mod4) occurs to even exponent in both m and n, so also in mn, and the primes p≡1(mod4) occurring in mn are exactly those occurring in one factor or the other, with the same exponents. Step 2.1 therefore factors the nonzero case as f(mn)=∏pe∥m, p≡1(4)(e+1)∏pe′∥n, p≡1(4)(e′+1)=f(m)f(n). Together with f(1)=1 from step 1.1, this is exactly the multiplicativity condition of Multiplicative arithmetic functions.

4.1step 1.1step 2.1step 3.1∎

Steps 1.1, 2.1, and 3.1 prove the stated prime-power values and multiplicativity.

Remarks

  • The load-bearing sourced input is Hackman Chapter K.III.1's exact formula for r2(N)/4. Steps 2.1 and 3.1 use that formula, read in the library's ordered-sign convention, to obtain both the prime-power values and the multiplicativity statement.

Depends on

Used by

Dependency tree · two levels

16 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