Im not really sure where to start
#prove perfect cubes help
leaden hemlock
dim pivotBOT
Im not really sure where to start
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hard summit
zoom in the problem statement plz
eternal pasture
Im not really sure where to start
use the prime factor decomposition
eternal pasture
zoom in the problem statement plz
"a, b, c integers such that c³ = a² b and gcd(a,b) = 1, show that a and b are perfect cubes"
hard summit
fundamental theorem of arithmetic
the decompositions can't have any primes in common by assumption, conclude the result
@leaden hemlock
leaden hemlock
use the prime factor decomposition
How with these variables? Do i just say a^2b|c^3? And then?
hard summit
what can you say about the exponents on the left hand side?
eternal pasture
write a = Πp p^vp(a) and b = Πp p^vp(b) their prime number decompositions. You must show that vp(a) and vp(b) are all multiple of 3
c³=a²b tells you that for all prime number p, 2 vp(a) + vp(b) is a multiple of 3, and the assumption on gcd(a,b) = 1 tells you that one of vp(a) or vp(b) is 0
conclude from this
leaden hemlock
write a = Πp p^vp(a) and b = Πp p^vp(b) their prime number decompositions. You m...
Do we have to show for when b is odd and b is even?
Since a is always even
eternal pasture
you don't need to make this distinction
simply consider a prime number p dividing a and show that its valuation is a multiple of 3
leaden hemlock
simply consider a prime number p dividing a and show that its valuation is a mul...
Oh, im just really bad at putting the proofs together so Im using practice questions. Im not quite sure how I would word it.
eternal pasture
|| Let p be a prime number. We show that vp(a) is a multiple of 3. We know from the identity c³ = a²b that 3vp(c) = 2vp(a) + vp(b). Now, a and b are coprime, hence vp(a) = 0 or vp(b)=0. In the case vp(a) =0, we're done because 0 is a multiple of 3. In the case vp(b) = 0, then 3 vp(c) = 2 vp(a). The numbers 3 and 2 are coprime, hence by Gauss Lemma, 3 must divide vp(a), so vp(a) is also a multiple of 3. ||
hard summit
communicate your argument to the reader
eternal pasture
I put a redaction here if you feel lost, but try to come up with a clear argument for the reader
leaden hemlock
I put a redaction here if you feel lost, but try to come up with a clear argumen...
does a^2b turn into 2vp(a)+vp(b) because the valuation of products has an additive identity? im a little confused about that
eternal pasture
does a^2b turn into 2vp(a)+vp(b) because the valuation of products has an additi...
yes
hasty tinselBOT
@leaden hemlock
sturdy girder
+close