Question

45* . Prove that if a, b, c are three different integers, then there exist in-

finitely many positive integers n such that a+n, b+n, c+n are pairwise rel-

atively prime. 

Answer 1

If the integers $a,b,c$ are distinct, then the number $h=(a-b)(a-c)(b-c)$ is different from zero. In case $h\ne\pm1$, let $q_1,\dots,q_n$ denote all prime $>3$ divisors of $h$.

If two or more among numbers $a,b,c$ are even, put $r=1$, otherwise put $r=0$. Clearly, at least two of the numbers $a+r$, $b+r$, $c+r$ will be odd. If $a,b,c$ give three diferent remainders upon dividing by $3$, put $r_0$. If two or more among $a,b,c$ give the same remainder $\rho$ upon dividing by $3$, put $r_0=1-\rho$. Clearly, at least two of the number $a+r_0, b+r_0, c+r_0$ will be not divisible by $3$.

Now, let $i$ denote one of the numbers $1,2,\dots, s$. Exists an integer $r_i$ such that none of the numbers $a+r_i, b+r_i, c+r_i$ is divisible by $q_i$. According to the Chinese remainder theorem, there exist infinitely many positive integer $n$ such that $n\equiv r\mod2$, $n=r_0\mod3$, and $n\equiv r_i\mod q_i$ for $i=1,2,\dots, s$.

We shall show that the numbers $a+n,b+n$ and $c+n$ are pairwise relatively prime. Suppose, for instance, that $(a+n_,b+n)>1$. Then  there would exist a prime $q$ such that $q|a+n$ and $q|b+n$, hence $q|a-b$, which implies $q|h$ and $h\ne\pm1$. Since $n\equiv r\mod 2$ and at least two of the numbers $a+n, b+n, c+n$ are odd, and we cannot have $q=2$. Next, since $n\equiv r_0\mod3$ at least two of the numbers $a+r_0,b+r_0,c+r_0$ are not divisible by 3, at least two of the numbers $a+n,b+n,c+n$ are not divisible by $3$, and we cannot have $q=3$.

Since $q|h$, in view of the definition of $h$, we have $q=q_i$ for a certain $i$ from the sequence $1,2,\dots, s$. However, in view of $n\equiv r_i\mod q_i$, or $n\equiv r_i\mod q$, and in view of the fact that none of the numbers $a+n,b+n,c+n$ is divisible by $q_i=q$, contrary to the assumption that $q|a+n$ and $q|b+n$. Thus, we provet that $(a+n,b+n)=1$. In a similat way we show that $(a+n,c+n)=1$, and $(b+n,c+n)=1$. Therefore the numbers $a+n,b+n$ and $c+n$ are pairwise relatively prime. Since there are infinitely many such numbers $n$, the proof is complete.