*wrong solution*

Nop, is 2016

Oh nvm claim may not be true..

$k=2016.$ If one erases all the $(x-4k),(x-(4k+1))$ in LHS and $(x-(4k+2)),(x-(4k+3))$ in RHS, then it can be proved that the remaining equation doesn't have real roots. It's clear that there is no root bigger than $2016$ or smaller than $1$. Suppose that there is a root $s$ in $(m,m+1)$ where $m$ is a positive integer. If $m\equiv 1,3\mod 4$, then one side would be less than zero while the other side wouldn't be. So $m\equiv 2,4\mod 4$. Let's first solve the case $m\equiv 2\mod 4$. Define $f(x)=\frac{(x+1.5)(x-1.5)}{(x+0.5)(x-0.5)}$, then we have $\prod_{i=0}^{503} f(s-4i-2.5)=1$。If we let $-0.5< a=s-(m+0.5) < 0.5$, then \[\prod_{i=-p}^{q}f(a-4i)=1\]where $p,q$ are non-negative integers (I'm too lazy to write down its exactly form). However, we have that \[\prod_{i=-p}^{-1} f(a-4i)=\prod_{i=1}^{p} f(a+4i) = \prod_{i=1}^{p} (1-\frac{2}{(a+4i)^2-0.25})\geq 1-\sum_{i=1}^{p} \frac{2}{(a+4i)^2-0.25}\]\[>1-\frac{1}{6}-\sum_{i=2}^{\infty}\frac{2}{(a+4i)^2-0.25}>1-\frac{1}{6}-\frac{1}{8}\sum_{i=1}^{\infty}i^{-2}=\frac{5}{6}-\frac{\pi ^2}{48}>\frac{1}{2}\]Similarly, $\prod_{i=1}^{q}f(a-4i)>\frac{1}{2}$. But $f(a)\geq 9$, so $\prod_{i=-p}^{q}f(a-4i)>\frac{9}{4}$, a contradiction. In the case that $m\equiv 4$, we can rewrite the equation as $\frac{(s-2)(s-2015)}{(s-1)(s-2016)}\prod_{i=-p}^{q}f(a-4i)=1$ But $s>4$, so $\frac{(s-2)(s-2015)}{(s-1)(s-2016)}>\frac{2\times 2013}{3\times 2014}$. Hence, $\frac{(s-2)(s-2015)}{(s-1)(s-2016)}\prod_{i=-p}^{q}f(a-4i)>\frac{2\times 2013\times 9}{3\times 2014\times 4}>1$, also a contradiction. Hope my solution is correct.....

v_Enhance wrote: I think the answer is indeed $k=2016$. Is this right? Seems to good to be true. Consider the $1008$ inequalities \begin{align*} (x-1)(x-4) &< (x-2)(x-3) \\ (x-5)(x-8) &< (x-6)(x-7) \\ (x-9)(x-12) &< (x-10)(x-11) \\ &\vdots \\ (x-2013)(x-2016) &< (x-2014)(x-2015). \end{align*}Notice that in all these inequalities, at most one of them has non-positive numbers in it, and we never have both zero. So it is OK to multiply them all together, and we get the conclusion. Thus we have a construction for $k = 2016$, which is clearly optimal since for $k < 2016$ the LHS and RHS must have two coinciding linear terms. If one of them is non-positive, you can't multiply them together. Take -3 < -2 and 2 < 300.

Indeed, I am idiot... I've deleted the obviously wrong solution. At least empirically I think the construction does indeed work, but the proof is certainly not the nonsense I wrote

I think an estimate of "how bad the gaps between positive terms can get" makes your construction work; specifically, the only cases that trouble you are when 4k+2 <x <4k+3. (In all other cases either sign is different or all of the inequality terms you mentioned are positive.) In such a case, we need a bound of something like prod(i=1~2016/4) (4n-1)*(4n)/(4n-2)(4n+1) < 9. It seems true (resembles wallis product?) and I think my estimate works, but not 100% confident.

Yep, that's exactly my idea too. I'm still trying to work out the details... not pretty, which is disappointing.

If x lies in the intervals (2,3) or (6,7) or (10,11) etc. we can perhaps change the inequalities to (x-4)(x-5) / ((x-3)(x-6)) > 1 (x-8)(x-9) / ((x-7)(x-10)) > 1 ... and also (x-1)(x-2016) < (x-2)(x-2015) --> (x-1)(x-2016) / ((x-2)(x-2015)) > 1 Then we can multiply these all together. --- We want to prove (x-2)(x-3)(x-6)(x-7)...(x-2014)(x-2015) > (x-1)(x-4)(x-5)(x-8)...(x-2013)(x-2016) for all x If x = 1, 4, 5, 8...2013, 2016, LHS > 0, RHS = 0 If x = 2, 3, 6, 7...2014, 2015, LHS = 0, RHS < 0 If x is in (1,2) or (3,4) or (5,6) etc. then LHS > 0, RHS < 0 If x is in (4,5) or (8,9) or (12, 13) etc. then multiply together (x-1)(x-4) < (x-2)(x-3) and (x-5)(x-8) < (x-6)(x-7) etc. as above in v_enhance's quote to obtain the desired result> If x is in (2,3) or (6,7) or (10,11) etc. see above

The construction 1, 4, 5, 8, 9, ..., 2012, 2013, 2016 and 2, 3, 6, 7, ..., 2014, 2015 works. You can use bounding to prove that the latter is always larger than the former (use $(x+1)(x+4) < (x+2)(x+3)$). Check for the intervals (- infinity, 1], [4k-2, 4k-1], [4k+1, 4k+2], and [2016, infinity).

OK, here's an olympiad-style proof. The bounding everyone mentioned is very natural, it's mostly annoying to find a good way to write it up. Consider the $1008$ inequalities \begin{align*} (x-1)(x-4) &< (x-2)(x-3) \\ (x-5)(x-8) &< (x-6)(x-7) \\ (x-9)(x-12) &< (x-10)(x-11) \\ &\vdots \\ (x-2013)(x-2016) &< (x-2014)(x-2015). \end{align*}Notice that in all these inequalities, at most one of them has non-positive numbers in it, and we never have both zero. If there is exactly one negative term among the $1008 \cdot 2 = 2016$ sides, it is on the left and we can multiply all together. Thus the only case that remains is if $x \in (4m-2, 4m-1)$ for some $m$, say the $m$th inequality. In that case, the two sides of that inequality differ by a factor of at least $9$. Now we claim \[ \prod_{k \ge 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < e. \]To see this, note that it's equivalent to prove \[ \sum_{k \ge 0} \log \left( 1 + \frac{2}{(4k+1)(4k+4)} \right) < 1. \]To this end, we use the deep fact that $\log(1+t) \le t$, and thus it follows from \[ \sum_{k \ge 0} \frac{1}{(4k+1)(4k+4)} < \frac12 \]which one can obtain for example by noticing it's less than $\frac14\frac{\pi^2}{6}$. This completes the proof, because then the factors being multiplied on by the positive inequalities to the left and right of our "bad" one are less than $e^2 < 9$, and we're okay.

Mathpanda and jereme: It's not that simple b/c when x is between 4k+2 and 4k+3, one of the inequalities have a different sign, you cannot multiply or divide. Explanation of bound above: Assume 4k+2 <x <4k+3. We want to show (x-1)(x-4).. (x-2013)(x-2016) < (x-2)(x-3).. (x-2014)(x-2015), both being negative. (x-4k-2)(x-4k-3)/(x-4k-1)(x-4k-4) is bounded below by 9. So we have (x-4k-2)(x-4k-3) <= 9* (x-4k-1)(x-4k-4). It suffices to show that all the other terms don't contribute a factor of 9 into the negative values. ...and we have a proof of it above.

qwerty414 wrote: Mathpanda and jereme: It's not that simple b/c when x is between 4k+2 and 4k+3, one of the inequalities have a different sign, you cannot multiply or divide. Explanation of bound above: Assume 4k+2 <x <4k+3. We want to show (x-1)(x-4).. (x-2013)(x-2016) < (x-2)(x-3).. (x-2014)(x-2015), both being negative. (x-4k-2)(x-4k-3)/(x-4k-1)(x-4k-4) is bounded below by 9. So we have (x-4k-2)(x-4k-3) <= 9* (x-4k-1)(x-4k-4). It suffices to show that all the other terms don't contribute a factor of 9 into the negative values. ...and we have a proof of it above. I can see where you are coming from, but I believed that I have adequately addressed the issue by altering the inequalities into a fractional form in my solution (where each inequality has positive LHS > 1) Then we can multiply these together to form ((x-1)(x-4)(x-5)(x-8)...)) / ((x-2)(x-3)(x-6)...)) > 1 which, as the denominator of the LHS is negative, can be multiplied out to form the desired result. The inequalities that I have used to deal with the 4k+2 and 4k+3 cases are different to those mentioned above in others' posts Excuse me if I am being ignorant but it would be great if you could verify this for me to check if it is correct or to point out the error

I have a question: what if this qurstion not purposed on 2016 but 20XX NOT divisible by 4?

jerememe wrote: qwerty414 wrote: Mathpanda and jereme: It's not that simple b/c when x is between 4k+2 and 4k+3, one of the inequalities have a different sign, you cannot multiply or divide. Explanation of bound above: Assume 4k+2 <x <4k+3. We want to show (x-1)(x-4).. (x-2013)(x-2016) < (x-2)(x-3).. (x-2014)(x-2015), both being negative. (x-4k-2)(x-4k-3)/(x-4k-1)(x-4k-4) is bounded below by 9. So we have (x-4k-2)(x-4k-3) <= 9* (x-4k-1)(x-4k-4). It suffices to show that all the other terms don't contribute a factor of 9 into the negative values. ...and we have a proof of it above. I can see where you are coming from, but I believed that I have adequately addressed the issue by altering the inequalities into a fractional form in my solution (where each inequality has positive LHS > 1) Then we can multiply these together to form ((x-1)(x-4)(x-5)(x-8)...)) / ((x-2)(x-3)(x-6)...)) > 1 which, as the denominator of the LHS is negative, can be multiplied out to form the desired result. The inequalities that I have used to deal with the 4k+2 and 4k+3 cases are different to those mentioned above in others' posts Excuse me if I am being ignorant but it would be great if you could verify this for me to check if it is correct or to point out the error I apologize. Been writing off my phone and misread your equations. It seems correct - and without using explicit bounds on values. Nice solution! (I may be mistaken)

Note that if we have the same $x-i$ on both sides we are doomed, so we need to erase at least $2016$ of them. So we want to prove that the answer is $2016$. This gives an idea of deleting $1008$ each. However, there is a problem. If we delete $1008$ each, then L.H.S, R.H.S are both monic 1008 degree polynomial. This forces us that L.H.S - R.H.S is a 1007 degree polynomial, which must have a real root. Darn. Unless... L.H.S and R.H.S both have the same $x^{1007}$ coefficient. This gives us an idea for partitioning $n \equiv 0, 1 \pmod{4}$ and $n \equiv 2, 3 \pmod{4}$ for our coefficients. Now to prove that this works... If $x$ is very small or very large, we have $(x-(4k+1))(x-(4k+4)) < (x-(4k+2))(x-(4k+3))$. (By very small or very large, I mean $x<1$ or $x>2016$) So we can just look at $[4k,4k+4]$ for each $k$. Cool. First, let's take a look at signs. Let $x \in [4k, 4k+4]$. Note that for all $u \not= k$, we have $0 \le (x-(4u+1))(x-(4u+4)) < (x-(4u+2))(x-(4u+3))$. If $x \in [4k+1, 4k+2] \cup [4k+3, 4k+4]$, then $(x-(4k+1))(x-(4k+4)) \le 0$ and $(x-(4u+2))(x-(4u+3)) \ge 0$ (Note, that even though I'm using $\ge$ and $\le 0$, since we deleted the common factors, we cannot have a zero in integers.) If $x \in [4k, 4k+1]$, then the inequality holds with $0 \le (x-(4k+1))(x-(4k+4)) < (x-(4k+2))(x-(4k+3))$ So we look at $x \in (4k+2, 4k+3)$. This is the main event because $(x-(4k+1))(x-(4k+4))$ and $(x-(4k+2))(x-(4k+3))$ are both negative. Say $x=4k+2+\epsilon$. We will look at the absolute value. This gives us $\frac{(x-(4k+1))(x-(4k+4))}{(x-(4k+2))(x-(4k+3))} = \frac{(1+\epsilon)(-2+\epsilon)}{\epsilon(\epsilon -1)}$ Discriminant shows that this value is at least $9$. Now look at the other magnitudes. We need to show that the multiple of $\frac{(x-(4u+1))(x-(4u+4))}{(x-(4u+2))(x-(4u+3))}$ where $u \not= k$, $0 \le u \le 503$ is more than $\frac{1}{9}$. So pretty much we can just prove $$\prod_{k \ge 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < 3 $$ That's done in the above. Take the natural log, use $\ln (1+x) \le x$, then use some bounds.

In the problematic case of $x\in(4m-2,4m-1)$, we can use the following instead, \begin{align*} (x-3)(x-6) &< (x-4)(x-5) \\ (x-7)(x-10) &< (x-8)(x-9) \\ &\vdots \\ (x-2011)(x-2014) &< (x-2012)(x-2013), \end{align*}where everything is positive. By combining this with the fact that $0<x-2<x-1$ and $0>x-2015>x-2016$, this completes the proof.

This might lead to a solution, but I'm not sure: Similarly to the following infinite product: $z\prod_{n\in\mathbb{Z}}\left (1- \frac{z^{2}}{n^{2}} \right )$ Which gives the sine function, Possibly the same (or a similar) infinite product for values of n which are 0 or 1 mod 4, give: $sin\left ( \frac{\pi}{2}n+\frac{\pi}{4} \right )-\frac{\sqrt{2}}{2}$ And for 2 or 3 mod 4: $-cos\left ( \frac{\pi}{2}n+\frac{\pi}{4} \right )+\frac{\sqrt{2}}{2}$ Since adding terms (the version of the question with a larger multiple of 4, other than 2016) only makes the problamatic case harder, considering the infinite product is enough. The two sine functions above are tangent in the problamatic intervals, which means there is a strict inequality for a finite version of the infinite product (equality in a finite amount of steps would make the inequality incorrect in any subsequent step) which is what we are asked to prove. I can't justify why the infinite product converges to a sine function, or something close to it which works similarly (maybe the ratio between the infinite products equal the ratio between the sine functions), but this seems like the natural guess for a function with zeroes in 0, 1 mod 4 over all integers. Any ideas?

Note that if both 2 sides have $x-i$ then the function has a root is $ x = i$ So we need to erase at least $2016$ of them. We will erase $x-(4k+2),x-(4k+3)$ at the $LHS$ and $x-4k, x- (4k+1)$ at the $RHS$ The function becomes : $(x-1)(x-4)(x-5)..(x-2016) = (x-2)(x-3)(x-6)...(x-2015)$ Note that $LHS$ and $RHS$ has the same sign if $x<1$, $4k < x < 4k+1$ , $4k+2<x<4k+3$ and $x>2016$ If $x<1$ then $x-i < 0$ $\forall$ $1 \geq i \geq 2016$. So, we consider the equation : $(1-x)(4-x)..(2016-x) = (2-x)(3-x)...(2016-x)$ We have : $0< (4k+1-x)(4k+4-x) < (4k+2-x)(4k+3-x)0$ $\forall$ $1 \leq k \leq 504$ then $LHS < RHS$ Prove similary for $x > 2016$ If $4k < x < 4k+1$ Suppose that $4<x<5$, we consider the equation : $(x-1)(x-4)(5-x)(8-x)...(2016-x) = (x-2)(x-3)(6-x)(7-x)....(2015-x)$ We have : $0< (x-1)(x-4) < (x-2)(x-3)$ and $0< (4k+1-x)(4k+4-x) <(4k+2-x)(4k+3-x)$ so $LHS < RHS$ Prove similary for $8<x<9,...$ so with $4k<x<4k+1$. the equation has no solutions We can prove similary with the case $4k+2 < x < 4k+3$. Then, the minimum value of $k$ is $2016$

We see that if number if deleted terms $ = x <2015$, then we would have at least one $x-p$ term on both sides and then $x=p$ satisfies the given equation. Hence, $x \geq 2016$. Now, we see that $\prod\limits_{k=0}^{503} (x-4k-1)(x-4k-4) = \prod\limits_{k=0}^{503} (x-4k-2)(x-4k-3)$ has no real solutions as $(x-4k-2)(x-4k-3) > (x-4k-1)(x-4k-4)$ implies that $x\in \bigcup_{k=1}^{503}(4k+2, 4k+3)$ so that $LHS = RHS < 0$ for the equality to have solutions and we see that this gives $$\left(\prod_{\substack{k\equiv 0,1\bmod{4}\\ 1\le k < x}}(x-k)\right) \left(\prod_{\substack{k\equiv 0,1\bmod{4}\\ x< k \le 2016}}(k-x)\right) >\left(\prod_{\substack{k\equiv 2,3\bmod{4}\\ 1\le k < x}}(x-k)\right) \left(\prod_{\substack{k\equiv 2,3\bmod{4}\\ x < k \le 2016}}(k-x)\right) $$after we take out the negatives from both the sides and we apply $(x-4k-2)(x-4k-3) > (x-4k-1)(x-4k-4)$ and $ (-x+4k+2)(-x+4k+3) > (-x+4k+1)(-x+4k+4)$ on every term except the first and last on both sides and these give that $(2016-x)(x-1) > (2015-x)(x-2)$, which is true implying our desired result. Hence, required number of deleted terms is $\boxed{x = 2016}$. Remark : I don't know how did I did that silly logical error so now I have corrected it. Also, I didn't know how to LaTeX that huge part so I took some help of some earlier solutions. Thanks for them to enable me to finish the solution, as I am not a regular English speaker.

I claim that our answer is $k = 2016$. Suppose we erase $\leq 2015$ factors. There will be $\geq 2017$ left, say $a$ and $b$ on the left and right hand side respectively. Since $a + b = 2017 > 2016$, there must be a factor $x - c$ on both sides of the equation, and this equation would then have root $c$, which is bad. Then, consider the following construction for $k = 2016$, where all LHS terms are $0, 1 \pmod 4$ and all RHS terms are $2, 3 \pmod 4$:\[(x - 1)(x - 4)\ldots(x - 2013)(x - 2016) = (x - 2)(x - 3)\ldots (x - 2014)(x - 2015)\]which we claim to have no real roots. Note that for all $x$ in intervals of the form $[4i + 1, 4i + 2]$ and $[4i + 3, 4i + 4]$, the two sides much have different signs, so no such roots can exist in these intervals. Furthermore, if $x \leq 1$ or $x \geq 2016$ both sides of the equality are positive but the inequality $(x - (4j + 1))(x - (4j + 4)) < (x - (4j + 2))(x - (4j + 3))$ holds for all $0 \leq j \leq 503$ so taking the product of all these yields LHS $<$ RHS. Hence it remains to consider $x$ in intervals of the form $[4i + 2, 4i + 3]$ and $[4i, 4i + 1]$. For these intervals $[4i + 2, 4i + 3]$ and $[4i, 4i + 1]$, we note the important identity that $(x - 4j)(x - (4j + 1)) > (x - (4j-1))(x - (4j + 2))$ for all $x \not \in$ intervals of the form $[4i + 2, 4i + 3]$ and $[4i, 4i + 1]$ and $j$. Along with noting that $0 < x - 2 < x - 1$ and $0 > x - 2015 > x - 2016$ we multiply and get LHS $\neq$ RHS, and we are done. We have considered all intervals covering all real numbers, and there are no real roots to this equation, as desired. $\blacksquare$

We claim the minimum is $2016$. First note $k\ge 2016$ since erasing any $2015$ (or less) factors will leave a common linear factor on both sides. We claim the following construction works, by having $p(x)$ have all the $0,1\mod 4$ roots and $q(x)$ the $2,3\mod 4$ roots: \[ \begin{array}{cccc} p(x) := (x-1)(x-4)&(x-5)(x-8)\ \cdots \ (x-2013)(x-2016) \\ q(x) := (x-2)(x-3)&(x-6)(x-7)\ \cdots \ (x-2014)(x-2015). \end{array} \]We need to check that $p(x)-q(x)=0$ has no real solutions. We claim $p(x)<q(x)$ for all $x\in \mathbb{R}$. Make cases on which of the following intervals $x$ lies in: \[ \begin{array}{c|cccc|cccc|c|c} (-\infty,1) & (1,2)&(2,3) & (3,4) & (4,5)&(5,6)&(6,7)&(7,8)&(8,9)&\cdots &(2016,\infty) \\ &a & b & c & d & a & b & c &d &\cdots \end{array} \]It suffices to use only open intervals since on the boundaries, exactly one of $p(x)$, $q(x)$ is zero, so their difference is nonzero. Case 1: $x\in (-\infty,1)$ or $x\in (2016,\infty)$. Note \begin{align*} (x-1)(x-4) \quad &< \quad (x-2)(x-3) \\ (x-5)(x-8) \quad &< \quad (x-6)(x-7) \\ (x-9)(x-12) \quad &< \quad (x-10)(x-11) \\ (x-13)(x-16) \quad &< \quad (x-14)(x-15) \\ &\ \vdots \end{align*}and that for $x<1$ and $x>2016$, both sides are positive for each row. Therefore, we can multiply the inequalities, and this implies $p(x) < q(x)$. Case 2: $x\in$ interval of type $a$ or type $c$. Then an odd number of terms of $p(x)$ are negative, and an even number of terms of $q(x)$ are negative. (For example, when $x\in (5,6)$, $3$ of $p(x)$ and $2$ of $q(x)$ are negative.) Hence $p(x)<0<q(x)$. Case 3: $x\in$ interval of type $b$ or type $d$. Note \begin{align*} x-1 \quad &> \quad x-2 \\ (x-4)(x-5) \quad &> \quad (x-3)(x-6) \\ (x-8)(x-9) \quad &> \quad (x-7)(x-10) \\ (x-12)(x-13) \quad &> \quad (x-11)(x-14) \\ &\ \vdots \\ (x-2012)(x-2013) \quad &> \quad (x-2011)(x-2014) \\ 2016-x \quad &> \quad 2015-x. \end{align*}The key is that for each row, both sides are positive. So multiplying gives $-p(x) > -q(x)$, i.e. $p(x)<q(x)$.

We claim the answer is $\boxed{k=2016}$. First part: Show that $k<2016$ doesn't work. Suppose that $k<2016$. Then there is some integer $1\le a\le2016$ such that $x-a$ is on both sides of the equation after erasing, which implies that $a$ is a real solution. Second part: Find a valid construction for $k=2016$. The construction is \[(x-1)(x-4)(x-5)(x-8)(x-9)(x-12)\cdots(x-2013)(x-2016)=(x-2)(x-3)(x-6)(x-7)(x-10)(x-11)\cdots (x-2014)(x-2015).\] Obviously there are no solutions for integers $1\le x\le 2016$. Claim: There are no solutions for $x<1$ or $x>2016$. Proof: We have \[(x-1)(x-4)=x^2-5x+4<(x-2)(x-3)=x^2-5x+6, (x-5)(x-8)<(x-6)(x-7),\]and so on until $(x-2013)(x-2016)<(x-2014)(x-2015)$. Note that $(x-1)(x-4), (x-2)(x-3), (x-5)(x-8),\ldots$ are all positive, which implies the right-hand side is strictly greater than the left-hand side. Now we only focus on $x$ such that $1<x<2016$. Claim: $x$ such that $4j+1<x<4j+2$ don't have any solutions, where $j$ is a non-negative integer. Proof: I will show that the sign (positive or negative) of the LHS is different from the sign of the RHS. We can do this by cancelling any two terms such that they have the same sign and are on different sides. So we can cancel everything except $x-(4j+1)$ and $x-(4j+2)$, which obviously have a different sign. Claim: $x$ such that $4j+3<x<4j+4$ don't have any solutions, where $j$ is a non-negative integer. We proceed similarly as above and we get that the LHS and RHS have different signs (by cancelling out everything except $x-(4j+3)$ and $x-(4j+4)$). Claim: $x$ such that $4j<x<4j+1$ don't have any solutions, where $j$ is a non-negative integer. Proof: Note that $(x-1)(x-4), (x-2)(x-3), (x-5)(x-8),\ldots$ are all positive (because for each of them, both of the factors are either $\ge x-4j$ or $\le x-(4j+1)$), which implies that LHS<RHS. Claim: $x$ such that $4j+2<x<4j+3$ don't have any solutions, where $j$ is a non-negative integer. This finishes the proof. Remove the factors $(x-1), (x-2), (x-2015), (x-2016)$ (we will put them back later). So the equation becomes \[(x-4)(x-5)(x-8)(x-9)(x-12)(x-13)\cdots(x-2012)(x-2013)=(x-3)(x-6)(x-7)(x-10)\cdots(x-2011)(x-2014).\] We have $(x-4)(x-5)>(x-3)(x-6), (x-8)(x-9)> (x-7)(x-10), \ldots, (x-2012)(x-2013)>(x-2011)(x-2014)$. All of these terms are positive (proof is similar to the above claim), which implies \[(x-4)(x-5)(x-8)(x-9)(x-12)(x-13)\cdots(x-2012)(x-2013)>(x-3)(x-6)(x-7)(x-10)\cdots(x-2011)(x-2014)>0.\] Now, $(x-1)>(x-2)>0$ and $(x-2016)<(x-2015)<0$, which implies $(x-1)(x-2016)<(x-2)(x-2015)<0$. So LHS=(small negative number) $\cdot$ (large positive number) = small negative number and RHS=(large negative number) $\cdot$ (small positive number)=large negative number, so LHS<RHS.

We claim the answer $k=2016.$ As the bound it's suffice to note, that no common factors should left in different sides. As an example, we erase $1008$ factors from each side in such way, that exactly polynomials $$P(x)=\prod_{i=1}^{504}(x-4i+3)(x-4i),$$$$Q(x)=\prod_{i=1}^{504}(x-4i+1)(x-4i+2)$$remains on different sides of equality. We claim, that $\forall x\in \mathbb{R}:P(x)\neq Q(x),$ which suffices. Case 1. $x\in (-\infty;1)\cup (1;+\infty)$ Notice that $0<(x-4k+3)(x-4k)<(x-4k+1)(x-4k+2)$ for $k=1,2,\dots,504.$ By multiplying we are done. Case 2. $x\in \left \{ 1,2,\dots,2016 \right \}$ In that case exactly one of $P(x),Q(x)$ is zero, hence the conclusion. In case $3$ we assume, that $x\in [1;2016]\backslash \mathbb{Z}.$ Here we consider two additional cases. Case 3.1 $\lfloor x \rfloor$ is odd. We easily obtain, that number of negative factors in $P(x),Q(x)$ are respectively odd and even, therefore $P(x)<0<Q(x).$ Case 3.2 $\lfloor x \rfloor$ is even. For $k=1,2,\dots,503$ we have $(x-4k)(x-4k+3)>(x-4k+1)(x-4k+2)>0.$ Moreover $(x-1)>(x-2)>0$ and $2016-x>2015-x>0,$ so multiplying altogether we obtain $-P(x)>-Q(x).$

First, let us prove the minimum $k$ is $k = 2016$. If we want to remove terms so $(x-1)(x-2)\cdots(x-2016)=(x-1)(x-2)\cdots (x-2016)$ will have no solution, there obviously must be no repeats on the 2 sides. If $k \le 2015$, there must be at least one overlapping term for the two sides, and all $k<2016$ don't work. For $k=2016$, we can take $\prod_{a=1}^{504}(x-(4a-3))(x-4a)$ for the LHS and $\prod_{a=1}^{504}(x-(4a-2))(x-(4a-1))$ for the RHS. If $x$ is not in the range of $[1, 2, 3, \dotsm, 2016]$, the RHS is always larger than the LHS If $x$ is in the range of $[1, 2, 3, \dotsm, 2016]$, at most 1 of $\prod_{a=1}^{504}(x-(4a-3))(x-4a)$ and $\prod_{a=1}^{504}(x-(4a-2))(x-(4a-1))$ is 0. If one of the 2 equations is zero, it is obvious it is not equal. Otherwise, the LHS is always smaller than the RHS. Therefore, the minimum $k$ that satisfies the requirements is $k=2016$

The answer is $2016$. Note that if we only remove $2015$ factors, then there exists a common factor on each side $x-a$, which implies that $a$ is a solution to the equation. $~$ Now, we'll show that \[\prod_{k=1}^{504}{(x-(4k-3))(x-4k)}=\prod_{k=1}^{504}{(x-(4k-2))(x-(4k-1))}\]has no real solutions. Clearly, there are no integer solutions between $1$ and $2016$. The key idea is that \[(x-(4k-3))(x-4k)=(x-(4k-2))(x-(4k-1))-2 < (x-(4k-2))(x-(4k-1)).\]Suppose $x<1$, $x>2016$, or $x\in (4k,4k+1)$ then every factor on the LHS is positive and less than every factor on the RHS, so the LHS is less than the RHS. If $x\in (4k-3,4k-2)\cup (4k-1,4k)$ for some integer $k\in \{1,\dots ,504\}$ then every factor on the LHS is positive except one, so it is negative. Every factor on the RHS is still positive, so it is positive. Again LHS $<$ RHS. $~$ It remains to see about $x\in (4k-2,4k-1)$. Note that $(x-(4k-3))(x-4k)$ and $(x-(4k-2))(x-(4k-1))$ are the only negative terms on each side. We have $(4k-x)(x-(4k-3))\ge 9 ((4k-1)-x)(x-(4k-2))$. $~$ Now, let $s=(x-(4(k-i)-2))$ be from $4i$ to $4i+1$ then \[\frac{(x-(4(k-i)-2))(x-(4(k-i)-1))(x-(4(k+i)-2))(x-(4(k+i)-1))}{(x-(4(k-i)-3))(x-4(k-i))(x-(4(k+i)-3))(x-4(k+i))}=\frac{s^2}{s^2-4}\]and it is easy to see from here that no amount of this being multiplied together with distinct positive $i$'s will ever exceed $9$.

We claim the answer is $2016$. If we remove at most $2015$ factors, we will have a factor repeated on both sides by Pidgeonhole, which clearly yields a real solution to the equation. For a construction, let the LHS only contain factors of the form $(x-a)$ where $a\equiv 1,4\pmod{4}$ and let the RHS contain the rest of the factors. Our equation is then \[\prod_{k=0}^{503}\left(x-(4k+1)\right)\left(x-(4k+4)\right)=\prod_{k=0}^{503}\left(x-(4k+2)\right)\left(x-(4k+3)\right)\]Notice \[\left(x-(4k+1)\right)\left(x-(4k+4)\right)+2=\left(x-(4k+2)\right)\left(x-(4k+3)\right)\quad(*)\]Hence, If $x<1$ or $x>2016$ one side is always larger than the other. If $x\in\{1,2,\dots,2016\}$ then one side is zero and the other side nonzero. If $x\in (4k+1,4k+2)\cup (4k+3,4k+4)$ for any $k\in \{0,1,\dots,503\}$ then one side is negative and the other is positive. We have one case remaining: suppose $x\in(4j+2,4j+3)$, ie $x=4j+2+\epsilon$ for $\epsilon\in(0,1)$, for some $j\in \{0,1,\dots,503\}$ is a solution to the equation. Consider \[P=\prod_{0\le k\le 503,k\neq j}\frac{\left(x-(4k+1)\right)\left(x-(4k+4)\right)}{\left(x-(4k+2)\right)\left(x-(4k+3)\right)}=\frac{\left(x-(4j+2)\right)\left(x-(4j+3)\right)}{\left(x-(4j+1)\right)\left(x-(4j+4)\right)}=\frac{\epsilon^2-\epsilon}{\epsilon^2-\epsilon-2}<\frac{1}{9}\]Let $f(k)$ be a term in $P$. Notice \[f(j-\ell)\ge \frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\]and \[f(j+\ell)\ge \frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\]by $(*)$. Notice $f$ increases as we get further from $j$. Thus, \begin{align*} P&=[f(j-1)f(j-2)f(j-3)\dots f(j-1008)][f(j+1)f(j+2)\dots f(j+1008)] \\&\ge\left(\prod_{\ell=1}^{1008}\frac{(4\ell)\cdot (4\ell-1)-2}{(4\ell)\cdot (4\ell-1)}\right)^2 \end{align*}It suffices to show \[\prod_{\ell=1}^{1008}\left(1+\frac{2}{(4\ell)(4\ell-1)-2}\right)<3\]in order to obtain a contradiction. We know $\ln(x+1)\le x$ so \begin{align*} \ln\left(\prod_{\ell=1}^{1008}\left(1+\frac{2}{(4\ell)(4\ell-1)-2}\right)\right)&\le\sum_{\ell=1}^{1008}\frac{2}{(4\ell)(4\ell-1)-2} \\&\le\sum_{\ell=1}^{1008}\frac{2}{(4\ell-1)^2} \\&=\frac{2}{9}+2\sum_{\ell=1}^{1008}\frac{1}{(4\ell+3)^2} \\&\le\frac{2}{9}+2\sum_{\ell=1}^{1008}\frac{1}{(4\ell)^2} \\&\le\frac{2}{9}+2\frac{1}{16}\cdot\frac{\pi^2}{6} \\&<\ln(3) \end{align*}as desired. $\square$

This proof seems shorter than others, can someone verify? If the two sides share a factor they share a 0 so it doesn't work, so $k\ge2016$. Now take LHS as $\prod_{k=1}^{504}(x-(4k-3))(x-4k)$ and RHS as $\prod_{k=1}^{504}(x-(4k-2))(x-(4k-1))$. Comparing each respective term it's easily computed that $(x-(4k-3))(x-4k)<(x-(4k-2))(x-(4k-1))$ (the key being for ANY x), so if both sides are negative LHS>RHS, if both sides are positive LHS<RHS, and both sides can't be zero, done. @below oh sorry ill try to get it fixed whoops my brain went out the window

signifance wrote: Comparing each respective term it's easily computed that $(x-(4k-3))(x-4k)<(x-(4k-2))(x-(4k-1))$ (the key being for ANY x), so if both sides are negative LHS>RHS, if both sides are positive LHS<RHS, and both sides can't be zero, done. $-2<-1$, $1<2$, but $-2\cdot 1=-1\cdot 2$.

Here is a terrible solution I was able to find. Answer: $2016$. Note that for every $k$, one of $(x-k)$ must be erased, otherwise, $k$ is the root of the remaining equation. Thus we have to erase at least $2016$ linear factor. Now for $0 \le k \le 503$, erase factors $(x - (4k+2)), (x - (4k+3))$ from RHS and erase factors $(x - (4k+1)), (x - (4k+4))$ from LHS. Then we will prove that the equation $\prod_{k=0}^{503}(x - (4k+1))(x - (4k+4)) = \prod_{k=0}^{503}(x - (4k+2))(x - (4k+3))$ has no real solution. If $x \not\in [1, 2016]$, then $(x - (4k+2))(x - (4k+3)) > (x - (4k+1))(x - (4k+4)) > 0$, hence $\prod_{k=0}^{503}(x - (4k+1))(x - (4k+4)) < \prod_{k=0}^{503}(x - (4k+2))(x - (4k+3))$. Thus we can assume $x \in [1, 2016]$. Also note that we may assume $x$ is not an integer. Now we'll split 3 cases. Case 1: $x \in [4s, 4s+1]$ for some $0 \le s \le 503$. In this case, we have $(x - (4s+2))(x - (4s+3)) > (x - (4s+1))(x - (4s+4)) > 0$ for all $0 \le k \le 503$, so $\prod_{k=0}^{503}(x - (4k+1))(x - (4k+4)) < \prod_{k=0}^{503}(x - (4k+2))(x - (4k+3))$. $\blacksquare$ Case 2: $x \in [4s+1, 4s+2]$ or $x \in [4s+3, 4s+4]$ for some $0 \le s \le 503$. Note that $(x - (4s+2))(x - (4s+3)) > 0 > (x - (4s+1))(x - (4s+4))$ and for all $s \neq k$, we have $(x - (4k+2))(x - (4k+3)) > (x - (4k+1))(x - (4k+4)) > 0$, thus $\prod_{k=0}^{503}(x - (4k+1))(x - (4k+4)) < 0 < \prod_{k=0}^{503}(x - (4k+2))(x - (4k+3))$. $\blacksquare$ Case 3: $x \in [4s+2, 4s+3]$ for some $0 \le s \le 503$. For the sake of contradiction, assume $\prod_{k=0}^{503}(x - (4k+1))(x - (4k+4)) = \prod_{k=0}^{503}(x - (4k+2))(x - (4k+3))$. Then we get $\frac{(x - (4s+1))(x - (4s+4))}{(x - (4s+2))(x - (4s+3))} = \prod_{k \neq s}(\frac{(x - (4s+2))(x - (4s+3))}{(x - (4s+1))(x - (4s+4))})$, or equivalently, $(1 + \frac{2}{(x - (4s+2))((4s+3) - x)}) = \prod_{k \neq s} (1 + \frac{2}{(x - (4k+1))(x - (4k+4))})$. Taking $\log$ from 2 sides and using the inequality $x \ge \log(x+1)$, we get $\log(1 + \frac{2}{(x - (4s+2))((4s+3) - x)}) < \sum_{k \neq s} (\frac{2}{(x - (4k+1))(x - (4k+4))})$. Combined with the fact that $9 \le 1 + \frac{2}{(x - (4s+2))((4s+3) - x)}$, we get $\log(3) < \sum_{k \neq s} (\frac{1}{(x - (4k+1))(x - (4k+4))})$. Note that $(\frac{1}{(x - (4k+1))(x - (4k+4))}) = \frac{1}{3}(\frac{1}{(x - (4k+4))} - \frac{1}{(x - (4k+1))})$, so $3\log(3) < \sum_{k \neq s} (\frac{1}{(x - (4k+4))} - \frac{1}{(x - (4k+1))}) < \sum_{k \neq s} (\frac{1}{(x - (4k+4))} - \frac{1}{(x - 4k)}) = \frac{1}{x - 4s} + \frac{1}{x - 2016} - \frac{1}{x} - \frac{1}{(x - (4s + 4))}) \le 3 < 3\log(3)$. Thus we get an evident contradiction. $\blacksquare$ Hence is all case, the equation $\prod_{k=0}^{503}(x - (4k+1))(x - (4k+4)) = \prod_{k=0}^{503}(x - (4k+2))(x - (4k+3))$ has no real solution, as needed. $\blacksquare$

The answer is $\boxed{2016}.$ We show that removing all with constant either $0,1 \pmod 4$ from the left side and $2,3 \pmod 4$ from the right side works. Note that for all $0 \leq k,$ we have that $$(x-(4k+1))(x-(4k+4))<(x-(4k+2))(x-(4k+3)).$$We can simply multiply all the inequalities of this form together whenever $\lfloor x \rfloor \not\equiv 2 \pmod 4.$ However, when $\lfloor x \rfloor \equiv 2 \pmod 4,$ we have that for exactly one inequality the left and right sides are both negative, where the left is maximum the product of $9$ and the right side. Therefore we show that the the $$\frac{\prod \text{all right sides}}{\prod \text{all left sides}}<9.$$Note that $$\prod_{k \geq 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}=\frac21 (\frac{3}{4} \cdot \frac{6}{5})\cdot (\frac{7}{8} \cdot \frac{10}{9}) <2.$$Note that we have two of these products because $$\frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}<\frac{(4k+2+n)(4k+3+n)}{(4k+1+n)(4k+4+n)},$$so because $2^2<9,$ we're done$.\blacksquare$

The answer is $\boxed{2016}.$ Obviously, if we remove less than $2016$ terms, there will still be common factors on both sides. For $2016,$ we make the following construction: $$(x-1)(x-4)(x-5)(x-8) \cdots = (x-2)(x-3)(x-6)(x-7) \cdots.$$If $x \ge 2016,$ or $x \le 0,$ then $(x-(4k+1))(x-(4k+4)) < (x-(4k+2))(x-(4k+3)),$ and all of these pairs of products are positive, so the left hand side is clear less than the right hand side. If $x \in (4k+1, 4k+4),$ but $x \not \in (4k+2, 4k+3),$ for some integer $0 \le k \le 503,$ then the LHS is less than $0,$ while the right hand side is not. Finally, if $x \in (4k+2, 4k+3),$ then $$\frac{(x-(4k+2))(x-(4k+3))}{(x-(4k+1))(x-(4k+4))} \le \frac{1}{9}.$$Now, we get that everything else is at most $$\log\left(\prod_{i=0}^{\max(2015-k, k-1)} \frac{(4i+2)(4i+3)}{(4i+1)(4i+4)}\right) = 2 \cdot \sum_{i=0}^{k} \log\left(1+\frac{2}{(4i+1)(4i+4)}\right)$$$$ \le \sum_{i=0}^{k} \frac{2}{(4i+1)(4i+4)} \le \frac{1}{8} \cdot \frac{\pi^{2}}{6} \le e^{2} < 9.$$Thus, this ratio cannot be $1.$

The answer is $k=2016$. Clearly we cannot have $k<2016$, otherwise at least one factor $x-k$ appears on both sides. To see that $k=2016$ is achievable, consider the polynomial \[P(x) = \prod_{i=0}^{503}(x-4i-1)(x-4i-4) - \prod_{i=0}^{503} (x-4i-2)(x-4i-3).\]We will prove that $P(x) > 0$ for all real numbers $x$. Let $Q(x)$ and $R(x)$ denote the two products, such that $P(x) = Q(x) - R(x)$. Claim: $Q(x) > R(x)$ when $x < 0$ or $x > 2016$. Proof: Multiply the inequalities $(x-4i-1)(x-4i-4) > (x-4i-2)(x-4i-3)$ for each $0 \leq i \leq 503$. We are guaranteed that both products are positive, so this yields the result. $\blacksquare$ Now, observe that for $x \in [4k+1, 4k+2]$ or $x \in [4k+3, 4k+4]$ for any integer $0 \leq k \leq 503$, $Q$ is nonnegative and $R$ is nonpositive, with equality not holding simultaneously. When $x \in (4k, 4k+1)$, we can multiply the two inequalities \begin{align*} \frac{(x-2)(x-3) \cdots (x-(4k-2))(x-(4k-3))}{(x-1)(x-4) \cdots (x-(4k-3))(x-4k)} &> 1 \\ \frac{(x-(4k+2))(x-(4k+3)) \cdots (x-2014)(x-2015)}{(x-(4k+1))(x-(4k+4)) \cdots (x-2013)(x-2016)} &> 1 \end{align*}where each inequality follows again by multiplying the same inequalities in the claim. The dirty part of the proof happens when $x \in (4k+2, 4k+3)$. Here, $Q(x) < 0$ and $R(x) < 0$, so we will prove that $\frac{Q(x)}{R(x)} < 1$ for these values of $x$. In particular, write \begin{align*} \log\left(\frac{(x-2)(x-3) \cdots (x-2014)(x-2015)}{(x-1)(x-4) \cdots (x-2013)(x-2016)}\right) &= \sum_{0 \leq i \leq 503, i \neq k} \log\left(\frac{(x-4i-2)(x-4i-3)}{(x-4i-1)(x-4i-4)}\right) + \log\left|(x-(4k+2))(x-(4k+3))\right| - \log\left|(x-(4k+1))(x-(4k+4))\right|\\ &< \sum_{0 \leq i \leq 503, i \neq k} \log\left(1+\frac 2{(x-4i-1)(x-4i-4)}\right) + \log|(x-(4k+2))(x-(4k+3))| \\ &< \sum_{0 \leq i \leq 503, i \neq k} \frac 2{(x-4i-1)(x-4i-4)} - \log 4 \\ &< 4 \sum_{i=0}^\infty \frac 1{(4i+2)(4i+5)} - \log 4 \\ &< 4 \left(\sum_{i=1}^\infty \frac 1{4i(4i+4)} + \frac 1{10}\right) - \log 4 \\ &< \frac{13}{20} - \log 4 < 0. \end{align*}Hence we get the result.