proving a function is injective and finding an infinite sequence $(a_k)$ so that $(f(a_k))$ is strictly increasing 
Let $N$ be the set of nonsquare positive integers. Let $f$ be a function defined on $N$ as follows: for an element $n\in N,$ consider all ways to choose positive integers $a_1,\cdots, a_k$ so that $n < a_1 < a_2<\cdots < a_k$ and $na_1\cdots a_k$ is a perfect square, and let $f(n)$ be the minimum value of $a_k$ over all such ways. For example, $f(2)=6$.


*

*Prove that $f$ is injective.

*Prove or disprove that $f$ that $f$ is strictly increasing.

*Prove that there exists an infinite sequence of nonsquare positive integers $a_k$ so that $(f(a_k))_{k=1}^\infty$ is strictly increasing.

Note that $f$ maps $N$ to $N$; for if not, then suppose $f(n)$ is a perfect square, and let $n < a_1 < \cdots < a_k = f(n)$ be so that $n a_1\cdots a_k$ is a perfect square and $a_k$ is minimal. By the minimality of $a_k, k = 1,$ as otherwise $na_1\cdots a_{k-1}$ is a perfect square. But then $n$ must be a perfect square, a contradiction. Also, any prime factor with an odd exponent in $a$ must divide at least one of $a_1,\cdots, a_k$, meaning that $f(a) \ge a + p,$ where $p$ is the largest prime factor in $a$ occurring to an odd exponent.
To show $f$ is injective, it might be useful to use a contradiction (e.g. contradict the minimality of $f(n)$) and assume $f(n) = f(m)$ for some $n < m$. but i'm not sure about the details. As for whether $f$ is strictly increasing or question 3, I'm not sure about those, though I think $f$ might not be strictly increasing.
 A: $1.$ If $f(a) = f(b), a < b,$ then let $a_1,\cdots, a_j$ and $b_1,\cdots, b_k = a_j$ be so that $a < a_1<\cdots < a_j$ and $b < b_1 < \cdots < b_k = a_j, a a_1\cdots a_j$ and $b b_1\cdots b_k$ are both perfect squares and $a_j = b_k$ is minimal. Then $ a a_1\cdots a_j b b_1\cdots b_k$ is a perfect square. Delete any factors that appear twice (observe that no $a_i$'s or $b_i$'s can be repeated, so these factors will be an $a_i$ and a $b_j$). The result is a perfect square that's a product of $a,b,$ and some elements all exceeding $a$ the largest of which is less than $b_k=a_j$. Sorting these elements gives a value $a_{k'}'$ less than $a_j$ so that there exist $a < a_{1}' < a_2'<\cdots < a_{k'}' < a_j$ with $aa_{1}'  a_2'\cdots a_{k'}'   $ equal to a perfect square, contradicting the minimality of $a_j$.
$2.$ $f$ is not increasing, since $f(11) = 22 < f(12) = 20$. To justify this, note that $f(11) \ge 22$ from above and $11\cdot 18 \cdot 22$ is a square. For $f(12)$, from above, $f(12) \ge 15$. Let $12 < a_1<\cdots< a_k$ be so that $P = 12 a_1\cdots a_k$ is a square and $a_k$ is minimal. $a_k > 15$ as $a_k = 15$ implies that none of $a_1\cdots a_k$ is divisible by $5,$ which occurs to an odd power in $P$. From above, at least one of the $a_i$'s must be divisible by $3$. $a_1 = 13$ implies $a_k\ge 26$ and $a_1 = 14$ implies $a_k \ge 21$. So $a_1 = 15,$ and $a_k \ge 20.$ $a_1 = 15, a_2 = 20, k=2$ works. If $20 > a_1 > 15$, then $a_1 \ge 18 \Rightarrow a_k \ge 21,$ so $a_1 < 18$. $a_1 = 17\Rightarrow a_k \ge 34$, and finally $a_1 = 16$ implies $a_k \ge 21$ (because there must be a number among $a_1,\cdots, a_k$ divisible by an odd power of $3$).
$3.$ To find the desired infinite sequence, it might be useful to obtain some sort of bounds on $f(p)$ when $p$ is prime. $f(p ) \ge 2p$. If there exists some number of the form $2 \cdot n^2$ between $p$ and $2p$ for any prime $p > 3$, then we are done as $f(p) = 2p$ for any prime $p > 3$. To prove this, it suffices to show that if $p > 2a^2,$ then $2p > 4a^2$. By Bertrand's postulate, there is a prime $p$ strictly between $a^2 $ and $2a^2$ for any $a > 1$. Then $f(p) = 2p$ as $p < 2a^2 < 2p$. In more detail, take a sequence of primes $(p_k)$ where $p_k$ is between $a_k^2$ and $2a_k^2$ for each k and $a_1= 2, a_{k+1} = \lceil \sqrt{2} a_k \rceil$. The sequence is clearly strictly increasing by induction as $ 2a_{k+1}^2 > p_{k+1} > a_{k+1}^2 > 2a_k^2 > p_k > a_k^2$.
