Re: Reference Needed

Bill Dubuque <wgd@xxxxxxxxxxxxxxxxxxxx> wrote:
Deep <deepkdeb@xxxxxxxxx> wrote: (*paraphrased*)

If A,B,C,D,a,b,c,d are integers > 1, with a,b,c,d squarefree,

(a,b) = (c,d) = 1, /N := sqrt(N)

then A /a + B /b = C /c + D /d

-> {a , b} = {c , d}

SIMPLER From A /a + B /b = C /c + D /d and its square, we infer that

the two sets {A /a , B /b} , {C /c , D /d} have equal sum s & product p

so the two sets are equal (being the set of roots of X^2 - s X + p). QED

Such uniqueness of sqrt sum decompositions holds more generally:

THEOREM If AA,BB,CC,DD in field F, CD not in F, 2 != 0 in F

then A + B = C + D -> {A, B} = {C, D}

PROOF Squaring AA+BB + 2AB = CC+DD + 2CD

-> AB = CD + f, f = (CC+DD-AA-BB)/2 in F

Squaring -> AABB = CCDD+ff + 2fCD -> CD in F if f != 0

Hence f = 0 -> AB = CD -> {A, B} = {C, D} have equal

sum s and product p so these two sets are equal (being the

roots of X^2 - s X + p, unique since F is a domain). QED

See also my prior posts on Besicovitch's theorem:
http://google.com/groups?selm=y8zu0njj1dt.fsf%40nestle.csail.mit.edu
http://google.com/groups/search?q=group:*math*+dubuque+Besicovitch+

--Bill Dubuque
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