The calculation is complicated so I'll just give a basic outline to solve this problem. define polynomial A(x)=(px+1)(px+2)...(px+p-1) define sum(1/(px+k)^2)=B(x)/C(x), C(x)=(px+1)^2(px+2)^2...(px+p-1)^2 sum(1/(px+k))=A'(x)/pA(x) differentiate the equality, sum(1/(px+k)^2)=-(A(x)A''(x)-A'(x)A'(x))/p^2A^2(x) Use the above identity and Wolstenholme's theorem to prove that both B(x) and C(x) are of the form a_0+p^3(a_1x+a_2x^2+...) where a_0,a_1,a_2,... are integers, then the results follows.

Hm...this problem is quite interesting - but it's straight forward if you are familiar with the general binomial theorem (and of course with working with fractions in modular artihmetic): For any complex number $ \alpha$ and any nonnegative integer $ k$ define the binomial coefficient $ \binom{\alpha}{k}$ as \[ \binom{\alpha}{k}= \frac1{k!}\prod_{j=0}^{k-1} (\alpha-j). \] Then the general binomial theorem states that for any $ x,y,\alpha\in\mathbb{C}$, \[ (x+y)^\alpha = \sum_{j=0}^{\infty}\binom{\alpha}{j} x^jy^{\alpha-j}. \] It is easy to check that $ \binom{-2}{j}=(-1)^j(j+1)$, so we have \begin{align*} f_p(x) &=\sum_{k=1}^{p-1} (px+k)^{-2} \\ &= \sum_{k=1}^{p-1}\sum_{j=0}^\infty (-1)^j(j+1)x^jp^jk^{-2-j} \\ &\equiv \sum_{k=1}^{p-1} \sum_{j=0}^{2} (-1)^j(j+1)x^jp^jk^{-2-j} \\ &= \sum_{j=0}^2 (-1)^j(j+1)x^jp^j\sum_{k=1}^{p-1}\frac{1}{k^{2+j}} \\ &= \left(\sum_{k=1}^{p-1}\frac{1}{k^2}\right) - 2xp\left(\sum_{k=1}^{p-1}\frac{1}{k^3}\right) + 3x^2p^2\left(\sum_{k=1}^{p-1}\frac{1}{k^4}\right) \pmod{p^3} \end{align*} Using Theorem 1 and 2 of http://www.mathlinks.ro/Forum/viewtopic.php?p=291959#291959, we however have \[ \sum_{k=1}^{p-1}\frac1{k^3} \equiv 0\pmod{p^2} \] and \[ \sum_{k=1}^{p-1}\frac{1}{k^4}\equiv \sum_{k=1}^{p-1} k^{p-5} \equiv 0\pmod p \] so it remains that \[ f_p(x)\equiv \sum_{k=1}^{p-1}\frac{1}{k^2} \pmod{p^3} \] which is independent from $ x$ so that the proposition is now obvious. Not a very nice solution and I doubt that this solution would yield full marks at a competition - does anybody have a more elementary one?

Yimin Ge wrote: $ \sum_{k = 1}^{p - 1}\sum_{j = 0}^\infty ( - 1)^j(j + 1)x^jp^jk^{ - 2 - j} \equiv \sum_{k = 1}^{p - 1} \sum_{j = 0}^{2} ( - 1)^j(j + 1)x^jp^jk^{ - 2 - j}$ I'm rather uncomfortable with this step. You're not proving a formal identity; is it clear that notions of convergence are valid $ \bmod p$? I think this step needs to be done in the $ p$-adic numbers. Edit: I'm definitely uncomfortable with this step. What you're assuming is that $ \sum_{k = 1}^{p - 1} \sum_{j = 3}^{\infty} ( - 1)^j (j + 1) x^j p^{j - 3} k^{ - 2 - j}$ is rational and nonzero $ \bmod p$, which doesn't follow from the premises of the problem. (Or perhaps I'm just not seeing an obvious way to do this.)

Here's a pretty efficient solution: We'll show that the fraction $f_p(x)$, when taken modulo $p^3$, is independent of $x$, whence the desired result will follow. First, a lemma.

Now, observe that \[f_p(x) = \sum\limits_{k = 1}^{\frac{p - 1}{2}} \left(\frac{1}{(px + k)^2} + \frac{1}{(px + p - k)^2}\right) \equiv \sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{p^2(2x^2 + 2x + 1) - 2kp + 2k^2}{(px + k)^2(px + p - k)^2} \pmod{p^3},\] where we have cross multiplied and reduced the numerator modulo $p^3.$ Reducing modulo $p$, \[\sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{1}{(px + k)^2(px + p - k)^2} \equiv \sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{1}{k^4} \equiv 0 \pmod{p},\] where the last step follows from the lemma. Hence, \[\sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{p^2(2x^2 + 2x + 1)}{(px + k)^2(px + p - k)^2} \equiv 0 \pmod{p^3},\] and therefore \[f_p(x) \equiv \sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{-2kp + 2k^2}{(px + k)^2(px + p - k)^2} \pmod{p^3}.\] It follows from cross-multiplication that \[f_p(x) - f_p(y) \equiv \sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{(-2kp + 2k^2)\left[(py + k)^2(py + p - k)^2 - (px + k)^2(px + p - k)^2\right]}{(px + k)^2(px + p - k)^2(py + k)^2(py + p - k)^2} \pmod{p^3}.\] For simplicity, denote by $\Delta$ the denominator of this quantity. Applying difference of squares, and reducing modulo $p^3$, this simplifes to \[\sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{(-2kp + 2k^2)\left[-2k^2p^2(y^2 + y - x^2 - x)\right]}{\Delta} \equiv \sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{-4k^4p^2(y^2 + y - x^2 - x)}{\Delta} \pmod{p^3}.\] Reducing modulo $p$, we find that \[\sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{-4k^4(y^2 + y - x^2 - x)}{\Delta} \equiv \sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{-4k^4(y^2 + y - x^2 - x)}{k^8} \equiv 4(x^2 + x - y^2 - y)\sum\limits_{k = 1}^{\frac{p - 1}{2}} \frac{1}{k^4} \equiv 0 \pmod{p}.\] Thus $f_p(x) - f_p(y) \equiv 0 \pmod{p^3},$ as desired. $\square$

Lemma:For $p>5$ we have $p^2\mid \sum_{i=1}^{p-1}\frac{1}{i^3}$ Proof: $\sum _{i=1}^{p-1} \frac{1}{i^3}=\sum_{i=1}^{\frac{p-1}{2}} \frac{1}{i^3}+\frac{1}{(p-i)^3}=p\sum_{i=1}^{\frac{p-1}{2}} \frac{(i^2+(p-i)^2-i(p-1)}{i^3(p-i)3}$ Now after we got $p$ out everything we care about is $\mathcal{O}(p)$ and so we work with : $\sum_{i=1}^{\frac{p-1}{2}} \frac{(i^2+(p-i)^2-i(p-i)}{i^3(p-i)3}=\sum_{i=1}^{\frac{p-1}{2}} \frac{3i^2}{-i^6}=\sum_{i=1}^{\frac{p-1}{2}} \frac{-3}{i^4}\equiv_p \frac{1}{2} \sum_{i=1}^{p-1} \frac{1}{i^4}=\frac{1}{2} \sum_{i=1}^{p-1} i^{p-5}\equiv_p 0$ $f_p(x)-f_p(y)=\sum_{k=1}^{p-1}\frac{p(x-y)(p(x+y)+2k)}{(px+k)^2(py+k)^2}=p(x-y)\sum_{k=1}^{p-1}\frac{(p(x+y)+2k)}{(px+k)^2(py+k)^2}$ And so from now on we care only about $\mathcal{O}(p^2)$ $$\sum_{k=1}^{p-1}\frac{(p(x+y)+2k)}{(px+k)^2(py+k)^2}\equiv_{p^2}\sum_{k=1}^{p-1}\frac{p(x+y)+2k}{(2pxk+k^2)(2pyk+k^2)}=\sum_{k=1}^{p-1}\frac{p(x+y)+2k}{k^4+2k^3p(x+y)}\equiv_{p^2} \sum_{k=1}^{p-1}\frac{p(x+y)+2k}{k^4+2k^3p(x+y)}-\frac{1}{2k^3}$$$$= \sum_{k=1}^{p-1}\frac{3k^4}{(k^4+2k^3p(x+y))k^3}=\sum_{k=1}^{p-1}\frac{3k}{(k^4+2k^3p(x+y))2}\sim \sum_{k=1}^{p-1} \frac{1}{k^2(cp+k)}$$Where $\sim$ is abbreviated by on constant $\frac{3}{2}$ and $c$ is set to be $2(x+y)$. $$\sum_{k=1}^{\frac{p-1}{2}} \frac{1}{k^2}\left (\frac{1}{cp+k}+\frac{1}{cp+p-k}\right )=p\sum_{k=1}^{\frac{p-1}{2}} \frac{1}{k^2}\frac{1}{(cp+k)((c+1)p-k)}$$But now we only care for $\mathcal{O}(p)$ and last sum becomes $\sum_{i=1}^{\frac{p-1}{2}} \frac{1}{i^4}$ which we already showed to be zero and so we're done.$\blacksquare$

USA TST 2002 D1 P2 wrote: Let $p>5$ be a prime number. For any integer $x$, define \[{f_p}(x) = \sum_{k=1}^{p-1} \frac{1}{(px+k)^2}\]Prove that for any pair of positive integers $x$, $y$, the numerator of $f_p(x) - f_p(y)$, when written as a fraction in lowest terms, is divisible by $p^3$. Claim: $f_p(x) \equiv \sum_{k=1}^{p-1} \frac{1}{k^2}$ Proof: Manipulating,$f_p(x) \equiv \sum_{k=1}^{p-1} \frac{1}{k^2} \cdot \frac{1}{(1-\frac{-px}{k})^2} \equiv \sum_{k=1}^{p-1} \frac{1}{k^2} (1-\frac{2px}{k}+\frac{3 p^2 x^2}{k^2}) \equiv \sum_{k=1}^{p-1} \frac{1}{k^2} \pmod {p^3}$ Hence $f_p(x)-f_p(y) \equiv \sum_{k=1}^{p-1} \frac{1}{k^2} -\sum_{k=1}^{p-1} \frac{1}{k^2} \equiv 0 \pmod p$

This is a soluiton which doesn't require any trickery, just some patience and careful manipulations. Lemma : If we want the numerator of $\sum \frac{p^ea+x}{p^eb+y}$, where the $x$ and $y$ may vary, to be divisible by $p^e$, then this is equivalent to proving that the numerator of $\sum \frac{x}{y}$ is divisble by $p^e$ Proof: Just multiply by all the denominators. I'll use this multiple times throughout the solution without explicitly noting that I use it. Also by $\equiv $ I always mean that the numerators are equivalent, not the sums! Step 1: Getting the first factor $p$. we can see that $$f_p(x)-f_p(y)=\sum_{i=1}^{p-1}\frac{1}{(px+i)^2}-\frac{1}{(py+i)^2}=\sum_{i=1}^{p-1}\frac{p(y-x)(p(x+y)+2i)}{(p^2xy+p(x+y)i+i^2)^2}$$We already got the first $p$ in the numerator and we can divide everything by $(y-x)$, so now we would want the numerator of the remaining sum to be divisible by $p^2$. If we put $x+y=t$ We have that the numerator of $$\sum_{i=1}^{p-1}\frac{pt+2i}{(p^2xy+pti+i^2)^2}\equiv \sum_{i=1}^{p-1}\frac{pt+2i}{(pti+i^2)^2}\equiv \sum_{i=1}^{p-1}\frac{pt+2i}{2pti^3+i^4} (mod p^2)$$ Step 2:Getting rid of a second factor $p$. To do this we are going to observe the following sum $$\frac{pt+2i}{2pti^3+i^4}+\frac{pt+2(p-i)}{2pt(p-i)^3+(p-i)^4}\equiv \frac{(p-i)^3(pt+2i)(2pt+(p-i))+i^3(pt+2(p-i))(2pt+i)}{(2pti^3+i^4)(2pt(p-i)^3+(p-i)^4)}\equiv $$$$\equiv \frac{((p-i)^3+i^3)2i(p-i)+3pti((p-i)^3-i^3)}{(2pti^3+i^4)(2pt(p-i)^3+(p-i)^4)}\equiv \frac{3pi^22i(p-i)+3pti(3pi^2-2i^3)}{(2pti^3+i^4)(2pt(p-i)^3+(p-i)^4)}(mod p^2)$$Notice how we have that $$2\sum_{i=1}^{p-1}\frac{pt+2i}{2pti^3+i^4} \equiv \sum_{i=1}^{p-1} \frac{3pi^22i(p-i)+3pti(3pi^2-2i^3)}{(2pti^3+i^4)(2pt(p-i)^3+(p-i)^4)} (mod p^2)$$Thus we get a second factor of $p$ and from now on we would want the numerator to be divisible only by $p$. Step 3:Getting rid of the final $p$. $$\sum_{i=1}^{p-1} \frac{3i^22i(p-i)+3ti(3pi^2-2i^3)}{(2pti^3+i^4)(2pt(p-i)^3+(p-i)^4}\equiv \sum_{i=1}^{p-1} \frac{3i^22i(-i)+3ti(-2i^3)}{i^4(p-i)^4)} \equiv \sum_{i=1}^{p-1} \frac{-6(t+1)i^4}{i^8} (mod p)$$This is equivalent to showing that $$\sum_{i=1}^{p-1} \frac{1}{i^4} \equiv 0(mod p)$$Let $g$ be a quadratic residue $mod$ $p$. We have $1\equiv g^{4(p-1)}(mod p)$ and the $i^4$ equivalent to $g^4;g^{4.2};\cdots ;g^{4(p-1)}$ in some order, thus $$\sum_{i=1}^{p-1} \frac{1}{i^4} \equiv 1+g^4+\cdots +g^{4(p-2)}=\frac{g^{4(p-1)}-1}{g^4-1}\equiv 0(mod p)$$because $p>5$, and thus we're finally done!

This is extremely simple with $p$-adic stuff. Notice the expansion \[\frac{1}{(px+k)^2} = \frac{1}{k^2}\cdot \frac{1}{\left(1-\left(-\frac{px}{k}\right)\right)^2} = \frac{1}{k^2}-\frac{2px}{k^3}+\frac{3p^2x^2}{k^4}-\cdots\]Since we are only interested in things at the $p^3$ level we get \[f_p(x) \equiv \sum_{k \in \mathbb{F}_p^{\star}}\left(\frac{1}{k^2}-\frac{2px}{k^3}+\frac{3p^2x^2}{k^4}\right) \equiv \sum_{k \in \mathbb{F}_p^{\star}}\frac{1}{k^2} \pmod {p^3}\]The last step is slightly nontrivial but if we use $\sum_{k=1}^{p-1}k^3 = \left(\frac{p(p-1)}{2}\right)^2$ and the similar summation for the fourth exponent we can be done. All in all note that $f_p(x) \pmod {p^3}$ has become independent of $x$ and so we can call it a day.

Here's an easier way to manipulate stuff. Note it suffices to show the result for $y=x-1$. Observe modulo $p^3$, \begin{align*} f_p(x-1) - f_p(x) &= \sum_{k=1}^{p-1} \frac{1}{px - k} - \sum_{k=1}^{p-1} \frac{1}{px + k} = \sum_{k=1}^{p-1} \left( \frac{1}{(px - k)^2} - \frac{1}{(px + k)^2} \right) \\ &= \sum_{k=1}^{p-1} \left( \frac{(px + k)^2 - (px - k)^2}{(px-k)^2(px + k)^2} \right) = \sum_{k=1}^{p-1} \left( \frac{4pxk}{(p^2x^2 - k^2)^2} \right) = 4px \textcolor{blue}{\sum_{k=1}^{p-1} \frac{k}{(p^2x^2 - k^2)^2 }} \end{align*}So it suffices to show twice the blue sum is $0$ modulo $p^2$. Indeed, \begin{align*} \sum_{k=1}^{p-1} \frac{2k}{(p^2x^2 - k^2)^2} &\equiv \sum_{k=1}^{p-1} \frac{2k}{(-k^2)^2} = \sum_{k=1}^{p-1} \frac{2}{k^3} = \sum_{k=1}^{p-1} \left( \frac{1}{k^3} + \frac{1}{(p-k)^3} \right) \equiv \sum_{k=1}^{p-1} \left( \frac{1}{k^3} + \frac{1}{3pk^2 - k^3} \right) \\ &= \sum_{k=1}^{p-1} \left( \frac{1}{k^3} + \frac{1}{3pk^2 - k^3} \right) = \sum_{k=1}^{p-1} \frac{3pk^2}{k^3(3pk^2 - k^3)} = 3p \textcolor{red}{\sum_{k=1}^{p-1} \frac{1}{k^3(3p-k) } } \end{align*}So it suffices to show that negative of the red sum is $0$ modulo $p$. Let $g$ be a primitive root modulo $p$. Using $\left\{ \frac{1}{1},\frac{1}{2},\ldots,\frac{1}{p-1} \right\}$ is a complete residue class modulo $p$ we get \begin{align*} \sum_{k=1}^{p-1} \frac{-1}{k^3(3p - k)} &\equiv \sum_{k=1}^{p-1} \frac{-1}{k^3(-k)} = \sum_{k=1}^{p-1} \frac{1}{k^4} = g^0 + g^4 + \cdots + g^{4(p-2)} = \frac{g^{4(p-1)} - 1}{g^4 - 1} \equiv 0 \end{align*}where in the last step we used $g^4 \not\equiv 1 \pmod{p}$ (as $p -1 \ne 4$). This completes the proof. $\blacksquare$

Let $S_j= \sum_{j=1}^{p-1} j^{-1}$. Observe that $$f_p(x) \equiv \sum_{k=1}^{p-1} (\frac{k^{-1}}{\frac{px}{k^2}+1})^2 \equiv \sum_{k=1}^{p-1} \frac{k^{-2}}1-\frac{2px}{k^2}+\frac{2p^2x^2}{k^4} \pmod{p^3}$$Hence, $f_p(x) \equiv S_2 - 2pxS_4+2p^2x^2S_6$. Observe that $f_p(x) \equiv f_p(y) \pmod{p^3} \iff 2p^2x^2S_6-2pxS_4 \equiv 2p^2y^2S_6-2pyS_4 \pmod{p^3} \iff 2pS_6(x-y)(x+y) \equiv 2S_4(x-y) \pmod{p^2}$. Since we want to make this true for all $x,y$, the condition is equivalent to $2pS_6(x+y) \equiv 2S_4 \pmod{p^2}$. Observe that $S_6 \equiv 1^6+2^6+...+(p-1)^6 \pmod{p}$, which is well known to be divisible by $p$, if $p-1 \not|6$, which is clearly true since $p>5$. Hence, $2pS_6(x+y) \equiv 0 \pmod{p^2}$, so we need to prove that $p^2|S_4$. This is not so hard! It is well known that $p|S_4$ (by the same argument used above). Then, we may take a primitive root $g \pmod{p}$ which is also a primitive root $\pmod{p^2}$. Therefore $g^k \equiv a \pmod{p^2} \implies g^k \equiv a \pmod{p} \implies S_4 \equiv 0 \pmod{p^2}$, so we are done. $\blacksquare$

Pair terms for $k=m$ and $k=p-m$. Let $a=px+\frac p2$ and $b=\left|\frac p2-m\right|$. Then \begin{align*} & \frac{1}{(px+m)^2}+\frac{1}{(px+p-m)^2} \\ =& \frac{(px+m)^2+(px+p-m)^2}{\Big((px+m)(px+p-m)\Big)^2} \\ =& \frac{2a^2+2b^2}{(a^2-b^2)^2} \\ \equiv& \frac{2a^2+2b^2}{b^4-2a^2b^2} \\ =& \frac{2a^2+2b^2}{b^2(b^2-2a^2)} \\ =& \frac{3}{b^2-2a^2}-\frac{1}{b^2}\pmod{p^3}. \end{align*}Let $-8a^2=-4p^2\left(2\left(x+\frac12\right)^2\right)=p^2C_x$. The second term is constant regardless of $x$, so it suffices to show the result for \[\sum_{k=1}^{(p-1)/2}\frac{1/2}{b^2-2a^2} =\sum_{k=1}^{p-1}\frac{1/4}{b^2-2a^2} =\sum_{k=1}^{p-1}\frac{1}{(p-2k)^2+p^2C_x}.\]So we want to show that \[p^3\,\Bigg\vert\,\sum_{k=1}^{p-1}\left(\frac{1}{(p-2k)^2+p^2C_x}-\frac{1}{(p-2k)^2+p^2C_y}\right).\]The sum is equal to \[\sum_{k=1}^{p-1}\left(\frac{p^2\left(C_y-C_x\right)}{\left((p-2k)^2+p^2C_x\right)\left((p-2k)^2+p^2C_y\right)}\right).\]Since the denominator is never a multiple of $p$, we can just instead prove that \[p\,\Bigg\vert\sum_{k=1}^{p-1}\left(\frac{1}{\left((p-2k)^2+p^2C_x\right)\left((p-2k)^2+p^2C_y\right)}\right).\]But since \[\sum_{k=1}^{p-1}\left(\frac{1}{\left((p-2k)^2+p^2C_x\right)\left((p-2k)^2+p^2C_y\right)}\right) \equiv \sum_{k=1}^{p-1}\left(\frac{1}{(2k)^4}\right) \equiv \sum_{k=1}^{p-1}k^4\pmod p,\]by the (very obscure) sum of $4$th powers formula we are done. $\blacksquare$

For $p$-adic reasons, we can treat $f_p(x)$ as a formal power series and cancel terms $\pmod{p^3}$. We write \[ f_p(x) = \sum_{k=1}^{p-1} \frac{1}{k^2} \sum_{j=0}^{\infty} (-1)^j \cdot (j+1) \left(\frac{px}{k}\right)^j. \]Reduce everything $\pmod{p^3}$ henceforth. Then by FLT \[ f_p(x) = \sum_{k=1}^{p-1} \frac{1}{k^2} \sum_{j=0}^{2} (-1)^j \cdot (j+1) (px\cdot k^{p-2})^j = \sum_{k=1}^{p-1} \frac{1}{k^2}. \]Hence $f_p(x)$ is constant modulo $p^3$, as desired.

We have \begin{align*} (px+k)\frac{1-\frac pkx+\frac{p^2}{k^2}x^2}k&\equiv\frac1k\left(px+k-\frac pkx(px+k)+\frac{p^2}{k^2}x^2(px+k)\right)\pmod{p^3}\\ &\equiv\frac1k\left(k+\frac{p^3}{k^2}x^3\right)\pmod{p^3}\\ &\equiv1\pmod{p^3} \end{align*} so the sum is equivalent to $$\sum_{k=1}^{p-1}\frac1{k^2}\left(1-\frac pkx+\frac{p^2}{k^2}x^2\right)^2\equiv\sum_{k=1}^{p-1}\frac1{k^2}\left(1-\frac{2p}kx+\frac{3p^2}{k^2}x^2\right)\pmod{p^3}.$$Therefore, it suffices to show $$\sum_{k=1}^{p-1}\frac{2p}{k^3}x$$and $$\sum_{k=1}^{p-1}\frac{3p^2}{k^4}x^2$$are constant modulo $p^3$. This follows from \begin{align*} \sum_{k=1}^{p-1}\frac2{k^3}&\equiv\sum_{k=1}^{p-1}\frac1{k^3}+\frac1{(p-k)^3}\pmod{p^2}\\ &=\sum_{k=1}^{p-1}\frac{3pk^2}{k^3(p-k)^3}\pmod{p^2}\\ &=3p\sum_{k=1}^{p-1}\frac1{k(p-k)^3}\pmod{p^2}\\ &=3p\sum_{k=1}^{p-1}\frac1{-k^4}\pmod{p^2} \end{align*}and \begin{align*} \sum_{k=1}^{p-1}\frac1{k^4}&\equiv\sum_{k=1}^{p-1}k^4\pmod p\\ &\equiv\sum_{i=0}^{p-2}g^{4i}\pmod p\\ &\equiv\frac{g^{4(p-1)}-1}{g^4-1}\pmod p\\ &\equiv0\pmod p. \end{align*}

Disappointed with how little considering the polynomial $P(x) = (x-1)\dots(x-(p-1))$ and variants modulo $p^5$ did, so I present a much more run-of-the-mill solution. We employ the following `remarkable' identity: \[ (px + k) \left(p^2x^2 - pxk + k^2 \right) \equiv k^3 \pmod{p^3}. \]Therefore, \[ \frac{1}{(px+k)^2} \equiv \left( \frac{p^2x^2}{k^3} - \frac{px}{k^2} + \frac{1}{k} \right)^2 \equiv \frac{3p^2x^2}{k^4} - \frac{2px}{k^3} + \frac{1}{k^2} \pmod{p^3}. \]From the given condition it is not hard to see that the sum of reciprocal $4$th powers is $0$ mod $p$. Consider the polynomial \[ P(x) = (x - 1)(x-2)\cdots(x-(p-1)) \equiv x^{p-1} - 1 \pmod{p}. \]The sum of reciprocal cubes, when taken mod $p^2$, is \[ -\frac{C_3}{C_0} \]where $C_3$ is the coefficient of the $x^3$ term. We will prove that $C_3 \equiv 0 \pmod{p^2}$. More particularly, consider \[ \providecommand{\FF}{\mathbb F} \sum_{(a,b,c) \in \FF_p^3} \frac{2}{abc} \equiv \sum_{(a,b,c) \in \FF_p^3} \frac{1}{ab} \left(\frac{1}{c} + \frac{1}{p-c} \right) \equiv \sum_{(a,b,c) \in \FF_p^3} \frac{1}{ab} \cdot \frac{p}{c^2} \]but this immediately gets the result from the fundamental theorem of symmetric polynomials and every single one of those terms being divisible by $p$. Here we use the fact that the degree is greater than $4$ and that $P(x) \equiv x^{p-1} - 1 \pmod{p}$. As such we are done.

Since the previous post I have found a slightly stranger way of dealing with the denominators. Let $P(x) = \prod_{n=1}^{p-1}(x-n) = \sum d_ix^i$, and as $p \ge 7$ we note that $p \mid d_1, \dots d_5$. It is not hard to prove $P(kp) \equiv P(p) \pmod{p^3}$ for any integer $k$. Define $Q_k(x) = \prod_{n=1}^{p-1} (x - (pk + n)^2)$. Observe that mod $p$ all of these are equivalent to $x^{p-1} - 2x^{\frac{p-1}{2}} + 1$. We consider $Q_k(x) - Q_{k+1}(x) = \sum c_{k,i} x^i$. Particularly if $p \nmid k$ we take modulo the polynomial evaluated at $k^2p^2$ modulo $p^5$, since we want to show $p^3$ divides $c_{k,1}$. This vanishes all of the $x^3$ or higher terms. In what follows, all products are taken from $n=1$ to $n=p-1$: \begin{align*} \prod((kp)^2 - (kp-n)^2) - \prod((kp)^2 - (kp+n)^2) &= \prod(2kp-n)(n) - \prod(2kp+n)(-n) \\ &= \prod(2kp-n)(n) - \prod(2kp+n)(n) - \prod(kp-n)^2 + \prod(kp+n)^2 \\ &= \prod(2kp-n)(n) - \prod(2kp+n)(n) - (\prod(kp-n) + \prod(-kp-n))(\prod(kp-n) - \prod(-kp-n)) \\ &= \prod(2kp-n)(n) - \prod(2kp+n)(n) - 4(d_2 k^2p^2 + (p-1)!)(d_3 k^3p^3 + c_1 kp). \end{align*}If we know that $p^2 \mid d_3$, we vanish that term too, and we end up with \begin{align*} \prod(2kp-n)(n) - \prod(2kp+n)(n) - 4(d_2 k^2p^2 + (p-1)!)(d_1 kp) &= \prod(2kp-n)(n) - \prod(2kp+n)(n) - 4(p-1)!d_1 kp \\ &= (p-1)! \left( \prod(2kp-n) - \prod(2kp+n) - 4c_1kp \right) \\ &= (p-1)! \left( 2d_3(2kp)^3 + 2d_12kp - 4d_1kp \right) \\ &= 0. \end{align*}Hence, modulo $p^3$ we have that the $x$ coefficients of $Q_k$ and $Q_{k+1}$ are the same, for $p \nmid k$. If $p \mid k$ we do the exact same argument again, but modulo a higher power of $p$; likewise for $p^2 \mid k$. The power of $p$ we need dividing $d_3$ actually gets weaker this way which is very nice. To finish, since $\frac{c_{k,1}}{P(kp)^2}$ is the value we are looking for, we just finish by noting that $c_{k,1} = c_{l,1}$ and $P(kp) = P(lp)$ modulo $p^3$, as desired.

We can rewrite $f_p(x)$ with generating functions as \[\sum_{k=1}^{p-1} \frac{1}{(1 + \frac pk x)} = \sum_{k=1}^{p-1} \frac{1}{k^2} \left(1 - 2 \cdot \frac pk \cdot x + 3 \cdot \frac{p^2}{k^2} \cdot x^2 + \ldots \right).\] Taking modulo $p^3$, we see that only the first three terms of the infinite summation matter, so \[f_p(x) \equiv \left(\sum_{k=1}^{p-1} k^{-2}\right) - 2xp \left(\sum_{k=1}^{p-1} k^{-3}\right) + 3x^2p^2\left(\sum_{k=1}^{p-1} k^{-4}\right) \equiv \sum_{k=1}^{p-1} k^{-2},\] as $p^2 \mid \sum k^{-3}$ and $p \mid \sum k^{-4}$. Hence $f_p(x)$ modulo $p^3$ only depends on $p$. $\blacksquare$