Prove that a b c is a cube of some integer. Given three integers a , b , and

seiyakou2005n1

seiyakou2005n1

Answered question

2022-05-24

Prove that a b c is a cube of some integer.
Given three integers a, b, and c such that a b + b c + c a is an integer too, prove that the product a b c is a cube.

Answer & Explanation

Amelie Douglas

Amelie Douglas

Beginner2022-05-25Added 8 answers

By dividing by a common factor if there is any, we can assume no prime number divides all of a , b , c. Our goal is to show that the exponent of any prime p in prime decomposition of a b c is divisible by 3.
Suppose p divides one of the numbers, WLOG let p a. Also, let p k be the greatest power of p dividing a.
Our assumption on sum of fractions being integer is just saying that a b c a 2 c + b 2 a + c 2 b. We see p a b c and hence p a 2 c + b 2 a + c 2 b and so p c 2 b thus p b or p c, but not both (as we assumed). Now we have two cases.
1) p b. Let p l be the greatest power of p dividing b. It's easy to see now that exponent of p in a b c is k + l
We have p k + l a b c, so p k + l a 2 c + b 2 a + c 2 b, hence p k + l a 2 c + c 2 b. The greatest power of p dividing c 2 b is p l and the greatest power of p dividing a 2 c is p 2 k . If these exponents were different, then the greatest power of p dividing sum of c 2 b and a 2 c would be min { 2 k , l } < k + l which is impossible. Hence l = 2 k and k + l = 3 k, as we wanted.
2) p c. This case is treated similarly - but now c takes role of a and a takes role of b. I will leave it for you to fill in details.
These two cases in conjuction give us desired conclusion.

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