3 typo

To expand on the answer in my comment, the proportion of integers $a$ for which $4a+1,4a+2,4a+3$ are all squarefree is $\prod_{p\not=2}(1-3/p^2)$, with the product taken over all odd primes $p$. As this product converges to a positive limit, there are infinitely many such $a$. A quick heuristic is to look modulo $p^2$. For odd prime $p$, precisely $p^2-3$ of the possible $p^2$ values of $a$ mod $p^2$ lead to $4a+1,4a+2,4a+3$ all being nonzero mod $p^2$, so this has probability $1-3/p^2$. Independence of mod $p^2$ arithmetic as $p$ runs through the primes suggests the claimed limit.

More precisely, if $\phi(n)$ is the number of positive integers $a\le n$ with $4a+1,4a+2,4a+3$ squarefree then $$\frac{\phi(n)}{n}\to\prod_{p\not=2}(1-3/p^2).\qquad\qquad{\rm(1)}$$ It's not too hard to turn this heuristic into a rigorous argument. If we let $\phi_N(n)$ denote the number of $a\le n$ such that none of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for a prime $p < N$, then the Chinese remainder theorem says that we get equality $$\frac{\phi_N(n)}{n}=\prod_{\substack{p\not=2,\\ p < N}}(1-3/p^2).\qquad\qquad{\rm(2)}$$ wherever $n$ is a multiple of $\prod_{\substack{p\not=2,\\ p < N}}p^2$ and, therefore, the error in (2) is of order $1/n$ for arbitrary $n$. It only needs to be shown that ignoring primes $p\ge N$ leads to an error which is vanishingly small as $N$ is made large. In fact, the number of $a \le n$ which are a multiple of $p^2$ is $\left\lfloor\frac{n}{p^2}\right\rfloor\le \frac{n}{p^2}$. The number of $a\le n$ which is a multiple of $p^2$ for some prime $p\ge N$ is bounded by $n\sum_{p\ge N}p^{-2}$. So, the proportion of $a\le n$ for which one of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for an odd prime $p\ge N$ is bounded by $\frac{4n+3}{n}\sum_{p\ge N}p^{-2}\sim (4+3/n)/(N\log 3\frac{4n+3}{n}\sum_{p\ge N}p^{-2}\sim3(4+3/n)/(N\log N)$. This means that $\phi_N(n)/n\to\phi(n)/n$ uniformly in $n$ as $N\to\infty$, and the limit (1) follows from approximating by $\phi_N$.

2 typo

To expand on the answer in my comment, the proportion of integers $a$ for which $4a+1,4a+2,4a+3$ are all squarefree is $\prod_{p\not=2}(1-3/p^2)$, with the product taken over all odd primes $p$. As this product converges to a positive limit, there are infinitely many such $a$. A quick heuristic is to look modulo $p^2$. For odd prime $p$, precisely $p^2-3$ of the possible $p^2$ values of $a$ mod $p^2$ lead to $4a+1,4a+2,4a+3$ all being nonzero mod $p^2$, so this has probability $1-3/p^2$. Independence of mod $p^2$ arithmetic as $p$ runs through the primes suggests the claimed limit.

More precisely, if $\phi(n)$ is the number of positive integers $a\le n$ with $4a+1,4a+2,4a+3$ squarefree then $$\frac{\phi(n)}{n}\to\prod_{p\not=2}(1-3/p^2).\qquad\qquad{\rm(1)}$$ It's not too hard to turn this heuristic into a rigorous argument. If we let $\phi_N(n)$ denote the number of $a\le n$ such that none of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for a prime $p < N$, then the Chinese remainder theorem says that we get equality $$\frac{\phi_N(n)}{n}=\prod_{\substack{p\not=2,\\ p < N}}(1-3/p^2).\qquad\qquad{\rm(2)}$$ wherever $N$ n$is a multiple of$\prod_{\substack{p\not=2,\\ p < N}}p^2$and, therefore, the error in (2) is of order$1/n$for arbitrary$n$. It only needs to be shown that ignoring primes$p\ge N$leads to an error which is vanishingly small as$N$is made large. In fact, the number of$a \le n$which are a multiple of$p^2$is$\left\lfloor\frac{n}{p^2}\right\rfloor\le \frac{n}{p^2}$. The number of$a\le N$n$ which is a multiple of $p^2$ for some prime $p\ge N$ is bounded by $n\sum_{p\ge N}p^{-2}$. So, the proportion of $a\le N$ n$for which one of$4a+1,4a+2,4a+3$is a multiple of$p^2$for an odd prime$p\ge N$is bounded by$\frac{4N+3}{N}\sum_{p\ge \frac{4n+3}{n}\sum_{p\ge N}p^{-2}\sim 4/\log N$(4+3/n)/(N\log N)$. This means that $\phi_N(n)/n\to\phi(n)/n$ uniformly in $n$ as $N\to\infty$, and the limit (1) follows from approximating by $\phi_N$.

1

To expand on the answer in my comment, the proportion of integers $a$ for which $4a+1,4a+2,4a+3$ are all squarefree is $\prod_{p\not=2}(1-3/p^2)$, with the product taken over all odd primes $p$. As this product converges to a positive limit, there are infinitely many such $a$. A quick heuristic is to look modulo $p^2$. For odd prime $p$, precisely $p^2-3$ of the possible $p^2$ values of $a$ mod $p^2$ lead to $4a+1,4a+2,4a+3$ all being nonzero mod $p^2$, so this has probability $1-3/p^2$. Independence of mod $p^2$ arithmetic as $p$ runs through the primes suggests the claimed limit.

More precisely, if $\phi(n)$ is the number of positive integers $a\le n$ with $4a+1,4a+2,4a+3$ squarefree then $$\frac{\phi(n)}{n}\to\prod_{p\not=2}(1-3/p^2).\qquad\qquad{\rm(1)}$$ It's not too hard to turn this heuristic into a rigorous argument. If we let $\phi_N(n)$ denote the number of $a\le n$ such that none of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for a prime $p < N$, then the Chinese remainder theorem says that we get equality $$\frac{\phi_N(n)}{n}=\prod_{\substack{p\not=2,\\ p < N}}(1-3/p^2).\qquad\qquad{\rm(2)}$$ wherever $N$ is a multiple of $\prod_{\substack{p\not=2,\\ p < N}}p^2$ and, therefore, the error in (2) is of order $1/n$ for arbitrary $n$. It only needs to be shown that ignoring primes $p\ge N$ leads to an error which is vanishingly small as $N$ is made large. In fact, the number of $a \le n$ which are a multiple of $p^2$ is $\left\lfloor\frac{n}{p^2}\right\rfloor\le \frac{n}{p^2}$. The number of $a\le N$ which is a multiple of $p^2$ for some prime $p\ge N$ is bounded by $n\sum_{p\ge N}p^{-2}$. So, the proportion of $a\le N$ for which one of $4a+1,4a+2,4a+3$ is a multiple of $p^2$ for an odd prime $p\ge N$ is bounded by $\frac{4N+3}{N}\sum_{p\ge N}p^{-2}\sim 4/\log N$. This means that $\phi_N(n)/n\to\phi(n)/n$ uniformly in $n$ as $N\to\infty$, and the limit (1) follows from approximating by $\phi_N$.