find all positive integer solutions for a3bc=24a^{3}bc = 24

Find all positive integer solutions for a3bc=24a^{3}bc = 24.

#Algebra

Note by Sagnik Seth
2 years, 4 months ago

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Comments

24=233=a3bc 24 = 2^3 \cdot 3 = a^3bc

Since the only prime factor whose exponent is at least 3 is 2, a a has to be 2. Then, there is a remaining factor of 3, but it can't be split up, so either b b or c c is 3 and the respective other is 1.

(a,b,c)=(2,1,3) (a,b,c) = (2,1,3)

(a,b,c)=(2,3,1) (a,b,c) = (2,3,1)

Now, there is also the possibility of a=1 a = 1 . In this case, b b has to be one of 24's divisors and c c is determined by that.

(a,b,c)=(1,1,24) (a,b,c) = (1,1,24)

(a,b,c)=(1,2,12) (a,b,c) = (1,2,12)

(a,b,c)=(1,3,8) (a,b,c) = (1,3,8)

(a,b,c)=(1,4,6) (a,b,c) = (1,4,6)

(a,b,c)=(1,6,4) (a,b,c) = (1,6,4)

(a,b,c)=(1,8,3) (a,b,c) = (1,8,3)

(a,b,c)=(1,12,2) (a,b,c) = (1,12,2)

(a,b,c)=(1,24,1) (a,b,c) = (1,24,1)

Henry U - 2 years, 4 months ago
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