Coprime

Let $n$ be a natural number. Prove that there are natural numbers $a_1, a_2, . . . , a_n$ and natural numbers $b_1, . . . , b_n$ such that the following conditions are satisfied.

(a) If $i \not= j$, then $a_i$ and $a_j$ are coprime.
(b) If $i \not= j$, then $b_i$ and $b_j$ are coprime.
(c) The numbers $a_i$ and $b_j$ are not coprime for all $i$ and for all $j$.

How is this possible?

Let $n$ be a natural number. Prove that there are natural numbers $a_1, a_2, . . . , a_n$ and natural numbers $b_1, . . . , b_n$ such that the following conditions are satisfied.

(a) If $i \not= j$, then $a_i$ and $a_j$ are coprime.
(b) If $i \not= j$, then $b_i$ and $b_j$ are coprime.
(c) The numbers $a_i$ and $b_j$ are not coprime for all $i$ and for all $j$.

How is this possible?

It's easy enough to create an example - e.g. 3,14 and 25 for the a set and 2,7,39 for the b set. This can then form the basis for a generalisation.

That doesn't satisfy the criteria; 3 & 2 are coprime contradicting (c)

It's possible.

Hint:

Consider prime factors, and their grouping in a_i,b_i.

Start with n=2 and see what you can come up with.

Sorry I still can't how it is possible. It is condition (c) that is making it difficult for me. If a number (a_i) is not coprime with say two numbers (b_i and b_j), how can those two numbers (b_i and b_j) be coprime?

2 is coprime with 3 and 9, but 3 and 9 are not coprime.

I am saying if u is in the set of numbers a_i and v,w are in the set of number b_i, then for the three conditions (a),(b), (c) to be satisfied, u is not coprime with v and w, but v and w are coprime.

I really can't see how this is possible.

---

From your reply, are you suggesting I have misinterpreted the question?

Sorry I still can't how it is possible. It is condition (c) that is making it difficult for me. If a number (a_i) is not coprime with say two numbers (b_i and b_j), how can those two numbers (b_i and b_j) be coprime?

As james22 has indicated: For a_i, the common factor with b_i is different to the common factor with b_j.

Further hint: Since sequence b has n entries, what does that tell you about the number of distinct prime factors of a_i?

If the common factor of a_i and each number in b is different, then a_i must have n distinct prime factors with each one being in common factor with a different number in b?

If the common factor of a_i and each number in b is different, then a_i must have n distinct prime factors with each one being in common factor with a different number in b?

And since a_i is arbitrary, they must all have at least n distinct prime factors (including the b_i).

If you've not already sussed it, have a go with the case n=2. I'd use $p_1,p_2,...$ rather than actual numbers.

I was thinking along the right lines, but was unfortunately shown a solution by my lecturer before I could work it out. To be honest I don't think I would have worked it out if I wasn't show the solution.

So for n=2,

$a_1=p_1 \cdot p_2 \ and \ a_2=p_3 \cdot p_4$

Then a possible case would be:

$b_1=p_1 \cdot p_3 \ and \ b_2 = p_2 \cdot \ p_4$

I was thinking of having the numbers a_i and b_i as primes rather than product of primes so I could never meet the criteria (c).