Prove that for a, b, c N |a, b, c|2/[a, b][b, c][c, a] = (a, b, c)2/(a, b)(b, c)

Question

Prove that for a, b, c N |a, b, c|2/[a, b][b, c][c, a] = (a, b, c)2/(a, b)(b, c)(c, a) Hint: Use the theorem 7.1 Suppose that a and b have prime power factorizations a = Pa11 Pann and b = pb11 pbnn. Then gcd(a, b) = (a, b) = Pmin(a1,b1) . . . Pmin(an, bn)n and lcm (a, b) = [a, b] = pmax(a1, b1)1 . . . Pmin(an, bn)n. Proof: If d|a and d|b then the prime power fractorization of d is Pd11 . . .Pdnn for some d1 such that d1 le a1, b1 for all i. Similaly, if a|c and b|c, then the prime power factorization of c is pc11 . . . pcnn for some c1 such that c1 ge a1, b1 for all i. hence we have our describe results. Note: Observed that max{x, y} = x + y = x + y for all x, y R. So, from the above lemma we can get the following If a and b are positive integers, then [a, b]. (a, b) = ab. Proof: Let aand b have prime power factorizations a =Pa11 . . . pann and b = Pb11 . . . Pbnn. Let Mj= max{aj, bj} and mj = min {aj, bj} for all j {1, . . ., n}. Then ,

Explanation / Answer

suppose prime factor for a, b, c is following

a=p1a1p2a2..........pnan

b=p1b1p2b2.........pnbn

c=p1c1p2c2.........pncn

first take left hand side

[a,b,c]2 /[a,b][b,c][c,a]

from theorem we know that

[a,b]=p1max(a1,b1)p2max(a2,b2).................pnmax(an,bn)

so left hand side will be

{p1max(a1,b1,c1)p2max(a2,b2,c2).................pnmax(an,bn,cn)}2/(p1max(a1,b1)p2max(a2,b2).................pnmax(an,bn))(p1max(b1,c1)p2max(b2,c2).................pnmax(bn,cn))(p1max(a1,c1)p2max(a2,c2).................pnmax(an,cn))

now suppose

ai>bi>ci for any 1<i<n

here we are assuming this case at the end you will realize that it will hold for all the cases

now see what happens to pi

suppose for example our i is 1

then a1>b1>c1

and look at p1

in the numerator ve (p1a1)2

while in denominator we will have (p1a1p1b1p1a1)

so we wiil have

(p1a1)2/ (p1a1p1b1p1a1)

=1/p1b1

now take right hand side

from theorem we can write it as

{p1min(a1,b1,c1)p2min(a2,b2,c2).................pnmin(an,bn,cn)}2/(p1min(a1,b1)p2min(a2,b2).................pnmin(an,bn))(p1min(b1,c1)p2min(b2,c2).................pnmin(bn,cn))(p1min(a1,c1)p2min(a2,c2).................pnmin(an,cn))

here condition we remain same otherwise we will be solving it wrongly

so

a1>b1>c1

and look at p1

in the neuminator we will have (p1c1)2

while in denominator we will have (p1b1p1c1p1c1)

so we wiil have

(p1c1)2/ (p1b1p1c1p1c1)

=1/p1b1

which is same as we have got in left hand side

do this for all the pi

and we will get same from both side for each i

so from this we can find that

both side are equal to

1/p1r1p2r2.....pnrn

where ri is the middle term among ai and bi and ci

in our example above it was b1 at i = 1

hence we can say that both sides are equal

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