4 integers smallest sum. $4$ different positive integers are placed on the vertices of a square. Each pair of numbers on opposite vertices are relatively prime, while each pair of numbers on consecutive vertices are not relatively prime. What is the least possible sum of the $4$ numbers?
My attempt: let’s call the numbers $a$, $b$, $c$ and $d$. $b$ must have at least $2$ factors, 1st  must be common with a and 2nd must be common with $c$. Similarly $d$ must have at least $2$ factors. None of the numbers must be prime. For example, $6$, $14$, $35$, $15$ would this work?
 A: If one of the corners is prime, then it divides its two neighbours by assumption that they are not coprime, so its two neighbours share a common divisor which is not allowed. Similarly, a corner cannot be the power of a prime. This means that each corner, when written as a product of primes, must have two distinct prime divisors. 
Consider two opposite corners $a$ and $b$. Write $a=\lambda_1 p_1p_2$ and $b=\mu_1 p_3p_4$ where the $p_i$ are prime and distinct. They may also be chosen so that $p_1,p_3|c$ and $p_2,p_4|d$, where $c$ and $d$ are the remaining pair of opposite corners. This allows us to write $c=\lambda_2 p_1p_3$ and $d=\mu_2p_2p_4$. It is easy to specify conditions on $\lambda _i$ and $\mu_i$ so that this is a solution but it's not necessary, since we can note that choosing $a'=p_1p_2$ etc. yields a solution with a smaller sum. 
So this is now boiled down to the case where $p_1,p_2,p_3,p_4$ are distinct and the corners are each a product of a pair of these primes. This means their sum is $a'+b'+c'+d'=p_1p_2+p_3p_4+p_1p_3+p_2p_4=(p_1+p_4)(p_2+p_3)$, and since any choice of 4 distinct primes will suffice, we can choose the smallest: $2,3,5,7$.
There are now just 3 possibilities to check! The rest I will leave to you.

 $(2+3)(5+7)=60$ 
 $(2+5)(3+7)=70$ 
 $(2+7)(3+5)=72$ 

This means that the smallest sum is

 60, with the numbers 10, 21, 15, 14

