Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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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.

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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 p1(mod4),

and

f(qa)={1,a even,0,a odd,for q3(mod4).

Facts & Assumptions

Given: A natural exponent a, a prime p1(mod4), a prime q3(mod4), and coprime positive integers m,n.

Proof

technique · direct
1.1

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

givenalgebra
2.1

Write a positive integer N as N=2αi=1rpieij=1sqjbj, where the pi1(mod4), the qj3(mod4), and all listed primes are distinct. If some bj is odd and N=x2+y2, write N=qj2k+1M with qjM. 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 qjx0 and qjy0, hence qj2qjM, 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.

step 1.1given
3.1

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 q3(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 q3(mod4) occurs to even exponent in both m and n, so also in mn, and the primes p1(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)=pem, p1(4)(e+1)pen, p1(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.

step 1.1step 2.1givenalgebra
4.1

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

step 1.1step 2.1step 3.1

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