I'll just sketch this, since it's not very hard. Note that $n^2 + k \mid n^4 \iff n^2 + k \mid k^2$ but since $1 \le k \le 2n$ we arrive at only three cases: $n^2 + k = k^2$, $2(n^2 + k) = k^2$, $3(n^2 + k) = k^2$. The first has no solutions with $k \ge 1$ since we can put $(k-1)^2 < n^2 < k^2$. The other two are Pell equations, and one can check that $n^2 \equiv 2 \pmod 7$ has no solutions at all for $k \pmod 7$ in either case. The assertion about infinitely many solutions then follows by using the Pell recursion, and taking modulo $7$.

v_Enhance wrote: The assertion about infinitely many solutions then follows by using the Pell recursion, and taking modulo $7$. Is this a well known result? For all solutions to a pell equation there are infinite many that has certain value mod 7, and mod other primes. @below thx

wwwrqnojcm wrote: v_Enhance wrote: The assertion about infinitely many solutions then follows by using the Pell recursion, and taking modulo $7$. Is this a well known result? For all solutions to a poll equation there are infinite many that has certain value mod 7. Solutions to the Pell's Equation can be written as a linear reccurence sequence. Now it's known that in a linear reccurence sequence there are either 0, 1 or infinitely many elements of a given residue modulo $m$. Every such a sequence has a period w.r.t. modulo $m$ and hence if you find $a_i$ and $a_j$ s.t. $a_i=a_j$, $i,j$ are minimal and $i<j$ you have the following properties: The elements $a_k; \forall k<i$ are the only elements of their respective residue modulo $m$, while there are infinitely many elements having same residue with $a_k; \forall k$ s.t. $i\le k<j$. If you're interested in this topic you can search modular periodicity of linear reccurence sequences on the Internet. It can be easily obtained that the reccurence relation for the solutions for $n$ in $2(k+n^2) = k^2$ is given by $a_{m+2} = 6a_{m+1} - a_m$. Working w.r.t modulo $7$ and initial solutions of $a_0=2, a_1=12$ the sequence will look like $\{2,5,0,2,5,0,....\}$. Hence by what I already mention there are infinitely many solutions for $n$ such that $n=7m, 7m+2, 7m+5$. The recurrence relation for the solutions for $n$ in $3(k+n^2) = k^2$ is given by $a_{m+2} = 14a_{m+1} - a_m$, with initial values of $a_0 = 6, a_1=72$. Working w.r.t. modulo $7$ the sequence will be: $\{6,0,1,6,0,1...\}$. Hence there are infinitely many solutions for $n$, s.t. $n=7m, 7m+1, 7m+6$.

Recurrence can be proved by analyzing pairs of $(a_i,a_{i+1})$

$n^2+k|n^4 \implies n^2+k|k^2$, so $k^2=k+n^2$ or $k^2=2(k+n^2)$ or $k^2=3(k+n^2)$. The first is impossible since it is just $(2k-1)^2-4n^2=1$. The second is just $k^2=2k+2n^2$, which is just $(k-1)^2-2n^2=1$. The solutions of this Pell Equation is $x_{i+1}=3x_i+4y_i$, $y_{i+1}=2x_i+3y_i$, with $x_1=3$ and $y_1=2$. Here, $n=y_i$. We can see that $y_i$ has $2,5,0$ repeated in $\pmod{7}$, so there are infinitely many $7k+2$, $7k+5$, and $7k$. The third is just $k^2=3k+3n^2$, which is just $(2k-3)^2-12n^2=9$. The solutions of this Pell Equation is $x_{i+1}=7x_i+12y_i$, $y_{i+1}=4x_i+7y_i$, with $x_1=7$ and $y_1=4$. Here, $n=\frac{3}{2}y_i$. Now $y_i$ has a period of $0,1,0,6$ so there are infinitely many $7k+1$, $7k+6$. If $n=7k \pm 3$, we have $n^2 \equiv 2 \pmod{5}$, leading to a square being $5$ in $\pmod{7}$, which is impossible.

Write as $n^2+a\mid n^4$ for some $1\le a\le 2n$. Note that this implies \[n^2+a\mid a^2 \le 4n^2.\]Now, one of the following three equalities must hold; \begin{align*} a^2&=n^2+a\\ a^2&=2n^2+2a\\ a^2&=3n^2+3a. \end{align*}Note that the first equality can't hold unless $a=1,n=0$ because it rearranges to $n^2=a(a-1)$, which can't have a solution for relatively prime reasons. Then, rearrange the other two equations as \begin{align*} a(a-2)&=2n^2\\ a(a-3)&=3n^2. \end{align*} Claim: We have $n\not\equiv 3,4\pmod{7}$. Solution: Suppose otherwise. If $n$ satisfies the first equation, we get \[a(a-2) = (a-1)^2-1 = 2n^2\equiv 4\pmod{7},\]which clearly has no solutions $a$. If $n$ satisfies the second equation, we get \[a(a-3)\equiv a(a-10)\equiv (a-5)^2-25\equiv (a-5)^2+3\equiv 3n^2\equiv 6\pmod{7},\]which clearly has no solutions $a$. $\fbox{}$ Now, we will show that the rest of the solutions are attainable. We first parametrize all solutions to $a(a-2)=2n^2$. Let $a-1=k$, then we have $k^2-2n^2=1$. Then, we have that for some integer $m$, we have \[k+n\sqrt{2}=(3+2\sqrt{2})^m.\]This implies that \[n=\frac{(3+2\sqrt{2})^m-(3-2\sqrt{2})^m}{2\sqrt{2}}.\]Taking the $\sqrt{2}\equiv 4\pmod{7}$ root, we get \[n\equiv \frac{11^m-(-5)^m}{8} \equiv 4^m-2^m.\]For $m\equiv 0\pmod{3}$ this gives $n\equiv 0\pmod{7}$, for $m\equiv 1\pmod{3}$ this gives $n\equiv 2\pmod{7}$, and for $m\equiv 2\pmod{3}$ this gives $n\equiv 5\pmod{7}$. As Pell equations generate infinitely many solutions $n$, it remains to solve the problem for $n\equiv 1,6\pmod{7}$. We now parametrize all solutions of $a(a-3)=3n^2$. It is clear that $3\mid a$, so we let $k=2a/3$ and rewrite our equation as \[3k/2(k/2-1) = n^2.\]Let $2n/3=m$, which we can do because $3\mid n$, to get \[(k-1)^2-1=k(k-2)=3m^2.\]Let $\ell = k-1$, now we see that $n$ is $3m/2$ for a solution $m$ of $\ell^2-3m^2=1$ given that $m$ is even. Similarly to before, we can write \[n=3m/2 = \frac{3}{2} \cdot \frac{(2+\sqrt{3})^b - (2-\sqrt{3})^b}{2\sqrt{3}} = \frac{\sqrt{3}(2+\sqrt{3})^b - \sqrt{3}(2-\sqrt{3})^b}{4}\]for some integer $b$ given that that expression is, in fact, an integer. Note that $(2+\sqrt{3})^b$ is $7+4\sqrt{3}$ for $b=2$, $26+15\sqrt{3}$ for $b=3$, $97+56\sqrt{3}$, etc. In general, $m$ is even iff $b$ is even; this can be proved by induction. Thus, we can take $b=2c$ and write \[n=3m/2 = \frac{\sqrt{3}(7+4\sqrt{3})^c - \sqrt{3}(7-4\sqrt{3})^c}{4}.\]Now, we define $\mathbb{F}_{49}$ to be the field extension of $\mathbb{F}_7$ containing $\sqrt{3}$. By the Frobenius Endomorphism, we can take $c=7^ed$ and see that \[n=3m/2 = \frac{\sqrt{3}(7+4\sqrt{3})^d - \sqrt{3}(7-4\sqrt{3})^d}{4}.\]Now, note that $7-4\sqrt{3}$ and $7+4\sqrt{3}$ are inverses of each other, so by taking $d=\pm 1$ we can get \[n = \frac{\sqrt{3}(7+4\sqrt{3}) - \sqrt{3}(7-4\sqrt{3})}{4} = \frac{24}{4} = 6,\]or that $n=1$. This gives that by taking $b=2\cdot 7^e$ or $b=-2\cdot 7^e$ (equivalently $b=12\cdot 7^e$) for large $e$ we can generate infinitely many solutions $n\equiv 1,6\pmod{7}$.

By difference of squares $(n^2+d)(n^2-d)=n^4-d^2$ so $k(n^2+d)=d^2$ for some $k$. Since $d \leq 2n$, $k \in \{1,2,3\}$ if $k=1$ $n^2=d^2-d$. If $d=n$ this is $n^2-n$ and if $d=n+1$ this is $n^2+n$, so no solutions here. if $k=2$ $n^2=\frac{d^2-2d}{2}$ and one can easily check that $n \equiv 3,4 \pmod{7}$ does not work. if $k=3$ $n^2=\frac{d^2-3d}{3}$ and the residues of the RHS are $0,4,4,0,6,1,6$. Obviously $n \equiv 3,4 \pmod{7}$ don't work. From the quadratic formula $$d=\frac{3+\sqrt{9+12n^2}}{2}$$It's clear that if $\sqrt{9+12n^2}$ is an integer the result follows. Note $n=3m$ so $x^2-12m^2=1$ for some $x \in \mathbb{Z}^+$. Notice the trivial solution $(7,2)$ and PELL we're done (upon noticing $n \equiv 0,1,2,5,6$ working is obvious from cycling).

We want to analyze $n$ for which there exists $\ell \in [1, 2n]$ satisfying\[n^2 + \ell \mid n^4 \iff n^2 + \ell \mid \ell^2.\]Write $\ell^2 = k(n^2 + \ell)$ for some positive integer $k$. Clearly $k < 4$ since else $kn^2 + k\ell = \ell^2 \leq 4n^2$ which is impossible. So either $\ell^2 - \ell = n^2, \ell^2 - 2\ell = 2n^2,$ or $\ell^2 - 3\ell = 3n^2$. If $\ell^2 - \ell = n^2$, we get a contradiction since $(\ell - 1)^2 < \ell^2 - \ell = n < \ell^2$. Hence this case is not possible. If $\ell^2 - 2\ell = n^2$ then we rewrite it as\[(\ell - 1)^2 - 2n^2 = 1.\]This is a Pell equation and indeed has infinitely many solutions $(\ell, n)$. Moreover, for these solutions generated by the Pell Equation, as $n$ gets large, $\tfrac{\ell}{n} \approx \sqrt{2}$ so we do not need to worry about the $\ell \in [1, 2n]$ condition. Thus, for any positive integer $e$, if\[(3 + 2\sqrt{2})^e = c + d\sqrt{2}\]then $n = d$ does yield a valid $\ell$. Let $n = d_i$ correspond the the $d$ value obtained when $e = i$. It is clear that $d_{i+2} = 6d_{i+1} - d_i$ holds from simply analyzing the binomial expansions. The initial values are $d_1 = 2$ and $d_2 = 12$, and we can see that the sequence $d_1, d_2, \ldots$ cycles through $2, 5, 0, 2, 5, 0,\ldots$ modulo $7$. So in this case, we generate infinitely many solutions for $n$ that are $0, 2, 5 \pmod 7$. If $\ell^2 - 3\ell = 3n^2$ then we rewrite it as\[(2\ell-3)^2 - 3(2n)^2 = 9\]which is indeed a Pell Equation with infinitely many solutions. We can see that as $n$ gets large $\tfrac{\ell}{n} \approx \sqrt{3}$ so we do not have to worry about the $\ell \in [1, 2n]$ condition. Thus, for any positive integer $e$, if\[3(7 + 4\sqrt{3})^e = c + d\sqrt{3}\]then $n = \tfrac{d}{2}$ does yield a valid $\ell$. Let $n = \tfrac{d_i}{2}$ correspond the the $d$ value obtained when $e = i$. It is clear that $d_{i+2} = 14d_{i+1} - d_i$ holds from simply analyzing the binomial expansions. The initial values are $d_1 = 12$ and $d_2 = 168$, and we we can see that the sequence $\tfrac{d_1}{2}, \tfrac{d_2}{2}, \ldots$ cycles through $6, 0, 1, 6, 0, 1, \ldots$ modulo $7$. So in this case, we generate infinitely many solutions for $n$ that are $0, 1, 6 \pmod 7$. After exhausting through all cases, we see that it is impossible to generate any $n$ that are $3, 4$ modulo $7$, and that it is possible to generate infinitely $n$ that are $0, 1, 2, 5, 6$ modulo $7$, so we are done. $\blacksquare$ Wow.

Suppose $n^2 + a\mid n^4$, where $1\le a\le 2n$. We have $n^2 + a \mid (n^2 + a)(n^2 - a) = n^4 - a^2$, so $n^2 + a\mid a^2$. Now, $4n^2\ge a^2$, so either $n^2 + a = a^2, 2(n^2 + a) = a^2,$ or $3(n^2 + a) = a^2$. If $n^2 + a = a^2$, then $a = (a+n) (a-n)$. This value must be positive, so $a>n$. But then $(a+n)(a-n)> a$, contradiction. If $2(n^2 +a ) = a^2$, then $a^2 - 2a \equiv \{0,1,3,6\}\pmod 7$, which means $n^2\equiv \{0,3,4,5\}\pmod 7$, so $n\equiv \{0,2,5\}\pmod 7$. The equation rearranges to $a^2 - 2n^2 = 2a$, which is equivalent to $(a-1)^2 - 2n^2 = 1$. The Pell Equation $x^2 - 2y^2 = 1$ has positive integer solutions of the form $(a,b)$, where $(3+2\sqrt{2})^k = x+y\sqrt{2}$, for any positive integer $k$. The $y$ coordinate of this solution satisfies the recursion $y_{i+2} = 6y_{i+1} - y_i$ for positive integers $i$, where $y_1 = 2, y_2 = 12$, which means it cycles $2,5,0,2,5,0,\ldots$ mod $7$, so infinitely many $n\equiv \{0,2,5\}\pmod 7$ satisfy the equation (clearly $1\le a\le 2n$ because otherwise we $(a-1)^2 - 2n^2$ would be at least $2n^2$, which exceeds $1$). If $3(n^2 + a) = a^2$, then $a^2 - 3a\equiv \{0,3,4,5\}\pmod 7$, so $n^2\equiv \{0,1,4,6\}\pmod $, which means $n\equiv \{0,1,2,5,6\}\pmod 7$. The equation now rearranges to $a^2 - 3n^2 = 3a$, which is equivalent to $(2a-3)^2 - 3(2n)^2 = 9$. The solutions to the Pell Equation $x^2 - 3y^2 = 9$ are of the form $3(7 + 4\sqrt{3})^k = x + y\sqrt{3}$ for any positive integer $k$. The $y$ coordinate of this solution satisfies the recursion $y_{i+2} = 14y_{i+1} - y_i$, where $y_1 = 12, y_2 = 168$. This sequence cycles $5,0,2,0,5,0,2,0\ldots, $ modulo $7$, hence dividing by $2$ gives $6,0,1,0,\ldots,$, so there are infinitely many $n\equiv \{0,1,6\}\pmod 7$ satisfying the equation (clearly $1\le a\le 2n$ because otherwise $(2a-3)^2 - 3(2n)^2$ would be too large for $n>1$). To see that no $n\equiv \{3,4\}\pmod 7$ work, notice that $n\equiv \{3,4\}\pmod 7$ was impossible for all three cases.

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Suppose we have $n^2+k \mid n^4$ for some $0 \leq k \leq 2n$. By the Euclidean Algorithm this becomes $n^2+k \mid k^2$. Since $n^2+k=k^2$ is impossible for positive $n$ and $k^2 \leq 4n^2$, it follows that we must either have $2n^2+2k=k^2$ or $3n^2+3k=k^2$. Now it is easy to check all posibilities $\pmod{7}$ to show that we cannot have $n \equiv 3,4 \pmod{7} \iff n^2 \equiv 2 \pmod{7}$, hence any $n$ of the form $7m+3,7m+4$ cannot be in $S$. It remains to show that infinitely many elements of $S$ are congruent to some other residue modulo $7$. We consider the two cases for the value of $k^2$ separately. Case 1: $2n^2+2k=k^2$. Then clearly $k$ is even, so let $k=2m$ for some $m$. The condition then becomes $n^2=2m(m-1)$. We restrict our search to the possibility that $2m=a^2$ and $m-1=b^2$. This works if and only if $a^2-2b^2=2$, which is a Pell-type equation that has solution $(a,b)=(2,1)$. By the theory of Pell Equations, if $$(2+\sqrt{2})(3+2\sqrt{2})^i \coloneqq x_i+y_i\sqrt{2},$$then $(a,b)=(x_i,y_i)$ is a solution. It is clear that we have the recursion $x_{i+1}=3x_i+4y_i$ and $y_{i+1}=2x_i+3y_i$. This recursion must be period $\pmod{7}$, since we repeat terms eventually by Pigeonhole. This, combined with the values of $x_iy_i$ for small $i$, implies that there exist exist infinitely $i$ such that $x_iy_i \equiv 0,2,5 \pmod{7}$, which correspond with $n \equiv 0,2,5 \pmod{7}$. Case 2: $3n^2+3k=k^2$. Then clearly $3 \mid k$, so let $k=2m$ for some $m$. The equality rewrites as $n^2=3m(m-1)$. Restricting again, this time to $3(m-1)=a^2,m=b^2$, we get the Pell-type equation $a^2-3b^2=-3$. Doing basically the same thing as in case 1 yields an infinite number of $(a,b)$ that correspond to $n \equiv 0,1,6 \pmod{7}$. Combining these two cases, it follows that $S$ has infinitely many elements congruent to one of $\{0,1,2,5,6\} \pmod{7}$, so we're done. $\blacksquare$

This is pretty routine with Pell equations. The condition boils down to $n^2 + k \mid k^2$ for some choice of $k$. On the other hand, $k \leq 2n$, so $n^2+k > \frac{k^2}4$. This means that $n^2 + k = \frac{k^2}2$ or $n^2+k = \frac{k^2}3$. Former Case: This case is equivalent to the Pell equation $$(k-1)^2-2n^2 = 2,$$which means that $k, n$ can be generated by the expansion of $$(3+2\sqrt 2)^n(2+\sqrt 2).$$In other words, solutions $(a_i, b_i)$ of this form satisfy the recursive relations \begin{align*} a_{i+1} &= 3a_i + 4b_i \\ b_{i+1} &= 2a_i + 3b_i. \end{align*}It is easy to check that there are infinitely many solutions that generate $n \equiv 0, 2, 5 \pmod 7$ in this case. Latter Case: This case yields the Pell equation $$(2k-3)^2-3(2n)^2 = 9.$$This again yields infinitely many solutions that generate $n \equiv 0, 1, 6 \pmod 7$. On the other hand, it is clear for QR reasons that $n \equiv 3, 4 \pmod 7$ fails by inspecting the two equations. This finishes the problem.

This is somewhat boring problem. Let $1 \le k \le 2n$ be a positive integer such that $n^2 + k \mid n^4$. Thus we have $n^2 + k \mid k^2$. Since $k \le 2n$ and $n^2 + k \neq k^2$, so we have $2(n^2 + k) = k^2$ or $3(n^2 + k) = k^2$. Now split it into 2 cases: Case 1: $2(n^2 + k) = k^2$. In this case, we have $(k-1)^2 - 2n^2 = 1$. It's not hard to see that the roots of Pell equation $x^2 - 2y^2 = 1$ is periodic modulo 7, so checking it gives $n \equiv 0, 2, 5 (7)$. Case 2: $3(n^2 + k) = k^2$. By taking mod 3, we see that $3 \mid k$ and $3 \mid n$. And note that $3(n^2 + k) = k^2$ implies $(\frac{2k}{3} - 1)^2 - 3(\frac{2n}{3})^2 = 1$. The roots of Pell equation $x^2 - 3y^2 = 1$ is periodic modulo 7, so checking the roots gives $n \equiv 0, 1, 2, 5, 6 (7)$. In either case, Pell equation has infinitely many solutions, we're done. $\blacksquare$

Let $1\leq k\leq n$ be the number for which $n^2+k\mid n^4$. Since $n^4 \equiv k^2\pmod{n^2+k}$, we have $n^2+k\mid k^2$. Since $k\geq1$, we have $n^2+k\leq k^2\Longrightarrow n+1\leq k\leq 2n$. More importantly, $\frac{k^2}{n^2+k}$ is an integer. Note that \[1<\frac{(n+1)^2}{n^2+n+1}\leq \frac{k^2}{n^2+k}\leq\frac{4n^2}{n^2+k}<4\]which means that $\frac{k^2}{n^2+k}\in\{2,3\}$. The Quadratic Residues $\pmod{7}$ are $\{0,1,4,2\}$. Case 1. $\frac{k^2}{n^2+k}=2$, which means $k^2-2k=2n^2\Longrightarrow (k-1)^2=2n^2+1$. If $n\equiv\pm3\pmod{7}$, then $(k-1)^2\equiv5\pmod{7}$ which is impossible. There are infinitely many solutions for $n\equiv0,\pm2\pmod{7}$ by Pell's Equation. (The recurrence relation taken $\pmod{7}$ is simply $t_{n+2}+t_{n+1}+t_n\equiv0\pmod{7}$ and the sequence alternates $\langle 0,2,-2,\ldots\rangle$. Case 2. $\frac{k^2}{n^2+k}=3$, which means $k^2-3k=3n^2\Longrightarrow (2k-3)^2=12n^2+9$. If $n\equiv\pm3\pmod{7}$, then $(2k-3)^2\equiv5\pmod{7}$ which is impossible. There are infinitely many solutions for $n\equiv0,\pm1,\pm2\pmod{7}$ by Pell's Equation. (The recurrence relation taken $\pmod{7}$ is simply $u_{n+2}+u_n\equiv0\pmod{7}$ and the sequence alternates $\langle 0, 6, 0, 1,\ldots\rangle$. The proof is complete. $\blacksquare$

We start using Euclid. Suppose the divisor is of the form $n^2+k$, where $1 \leq k \leq 2n$: \[n^2+k \mid n^4 - (n^4-k^2) = k^2 \in \{1^2,2^2,\ldots,4n^2\} \implies x := \frac{k^2}{n^2+k} \in \{1,2,3\}.\] We have no solutions for $x=1$, but $x=2,3$ give \[k = 1+\sqrt{2n^2+1}, \quad \frac{3+\sqrt{12n^2+3}}{2}.\] Both solutions are clearly in bounds as $n \rightarrow \infty$, and notice $n \equiv 3,4 \pmod 7$ make each discriminant congruent to $5 \pmod 7$, which is not a quadratic residue, as desired. For the rest of the residues, we set each discriminant to $m^2$ and apply the linear recurrence generated by Pell to get infinitely many solutions for each class. $\blacksquare$

Suppose $n^2+k\mid n^4.$ By euclidean alg, $n^2+k\mid k^2.$ Since $k\le 2n$ we have $k^2\le 4n^2$ so we have three cases now. In the first case $n^2+k=k^2$ or $n^2=k(k-1)$ which has no solutions since $k,k-1$ are relatively prime and can't both be squares. In the second case $2(n^2+k)=k^2$ which is $k^2-2k-2n^2=0$ or $(k-1)^2-2n^2=1.$ The solutions to this pell equation are known, with the smallest solution $(k-1,n)=(3,2)$ and all solutions derivable from $(3+2\sqrt2)^m.$ Now we can get a recurrence $(a+b\sqrt2)(3+2\sqrt2)=(3a+4b+(2a+3b)\sqrt2).$ Now taking $\pmod 7$ we get that $(k-1,n)$ cycles $(3,2)\to(3,5)\to(1,0)\to(3,2)$ so this case gives infinitely many solutions to $n\equiv 0,2,5\pmod 7.$ In the third case $3(n^2+k)=k^2,$ so $k=3k'$ gives $n^2+3k'=3k'^2,$ and $n=3n'$ gives $3n'^2+k'=k'^2.$ Then $12n'^2=4k'^2-4k',$ so $(2k'-1)^2-12n'^2=1.$ We get the smallest solution is $(2k'-1,n')=(7,2)$ and the recurrence $(a+b\sqrt{12})(7+2\sqrt{12})=(7a+24b+(2a+7b)\sqrt{12})$ and $\pmod 7$ we get that $(2k'-1,n')$ cycles $(0,2)\to(6,0)\to(0,5)\to(1,0)\to(0,2)$ so this case gives infinitely many $n'\equiv 0,2,5\pmod 7,$ so $n=3n'\equiv 0,1,6\pmod 7.$ These are all possible cases since $4(n^2+k)>4n^2\ge k^2$ so we have characterized all possible $n,$ giving infinitely many $n\equiv 0,1,2,5,6\pmod 7$ and no $n\equiv 3,4\pmod 7,$ done.

We have $n^2 + k \mid n^4$, where $1 \leq k \leq 2n$. Also $n^2 + k \mid n^4 - (n^2 + k).n^2 + (n^2 + k).k$ $\Rightarrow$ $n^2 + k \mid k^2$. Now we have that $(n^2 + k).p = k^2$ and since $k \leq 2n$ we get that $(n^2 + k).p \leq 4n^2$ $\Rightarrow$ $p < 4$. When p = 1 we get the case $n^2 + k = k^2$ aka $n^2 = k(k-1)$ which has no solutions since gcd(k,k-1) = 1 and k and k-1 can't both be squares. Case 1: p = 2 We have $2(n^2 + k) = k^2$, which is equivalent to $(k - 1)^2 - 2n^2 = 1$ which is a Pell equation. With the smallest solution $(k-1,n) = (3,2)$ we get all solutions from $(3+2\sqrt2)^m$. Taking $\pmod 7$ we get that $(k-1,n)$ cycles $(3,2)\to(3,5)\to(1,0)\to(3,2)$ so from this case we get infinitely many solutions to $n\equiv 0,2,5\pmod 7$. Case 2: p = 3 We have $3(n^2 + k) = k^2$, which is equivalent to $(\frac{2k}{3} - 1)^2 - 3(\frac{2n}{3})^2 = 1$, which is a Pell equation. This is ok to do since $3 \mid k$ and $3 \mid n^2 + k$ $\Rightarrow$ $3 \mid n$. We have the Pell equation $x^2 - 3y^2 = 1$ and the smallest solution is $(2k'-1,n') = (7,2)$ and $\pmod 7$ we get that $(2k'-1,n')$ cycles $(0,2)\to(6,0)\to(0,5)\to(1,0)\to(0,2)$ so this case gives infinitely many $n'\equiv 0,2,5\pmod 7,$ so $n=3n'\equiv 0,1,6\pmod 7$. Now we have found all n that work and they are infinitely many $n \equiv 0,1,2,5,6 \pmod 7$ and no $n \equiv 3,4 \pmod 7$ $\Rightarrow$ we are ready.

Time-consuming problem, but each step is standard and well motivated

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