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+1 w
such that for every polynomial 
with integer coefficients and for every integer 
the integer![\[p^{(j)}(k)=\left. \frac{d^j}{dx^j}p(x) \right|_{x=k}\]](http://latex.artofproblemsolving.com/a/2/7/a2793d7f46345bcba4ccaa3082cc47afcd35074b.png)
(the -th derivative of at 
) is divisible by 
has always integral solutions for 
and 
whenever 
is a positive integer.
be an infinite, nonconstant sequence of rational numbers, meaning it is not the case that 
Suppose that 
is also an infinite, nonconstant sequence of rational numbers with the property that 
is an integer for all 
and . Prove that there exists a rational number 
such that 
and 
are integers for all and .
such that 
.
then using Mean Value Theorem, there exist an such that 
case, suppose 
and assume 
on the whole interval 
. Then by the mean value theorem, there is some such that 
so 
. That by itself isn't a contradiction, but we can do something similar to get a contradiction. The intuitive idea is that since we're right on the "edge" of a contradiction, we have no room to move the graph of 
. If 
and , that forces 
on the interval ![$[0,1/2]$](http://latex.artofproblemsolving.com/8/2/4/8240ac9beaadf3f2654c3ba09a3987a7305539c5.png)
. Moving the graph up or down at any point makes the slope larger than 
somewhere. Similarly, 
on the interval ![$[1/2,1]$](http://latex.artofproblemsolving.com/5/5/8/5581510e950d8ceb9bcbcfd0dc988d9c4538c684.png)
. for every ![$x\in [0,1/2]$](http://latex.artofproblemsolving.com/0/f/d/0fd16eeab1cb0c53e329c87cea809db55ea5f207.png)
. On any interval ![$[0,a]$](http://latex.artofproblemsolving.com/6/8/b/68bacdc4b96052586744d16169ccd8c3233ed09f.png)
with 
, we have 
for some 
, depending on . Since we're assuming 
is always bounded by , this means that 
so 
for all ![$a\in[0,1/2]$](http://latex.artofproblemsolving.com/e/0/c/e0c5302f45677b07a169c44aa1085cf1dace8463.png)
.
, then we could use the mean value theorem on the interval ![$[a,1/2]$](http://latex.artofproblemsolving.com/f/7/c/f7c796537cc023b8e3d0dd550cf1c3bdfc0278d4.png)
to get 
But the left side is larger than , and the right side is at most , a contradiction. on the interval , so 
However, this function is not differentiable at 
, so we are done.
:
Assume towards a contradiction that 
.
and 
:![\[\usepackage{mathtools}
\tfrac {1-f(x)}{\frac 12-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\tfrac 12\right):f(x)\ \ge\ 2x\tag{1}\]](http://latex.artofproblemsolving.com/4/6/d/46d07da0b23e15de91ca1912093a406842150f21.png)
LMVT on and 
:![\[\usepackage{mathtools}
\tfrac {f(x)-1}{x-\frac 12}\ge -2\xRightarrow{\hspace{40pt}}\left(\forall \tfrac 12<x< 1\right):f(x)\ \ge\ 2-2x\tag{2}\]](http://latex.artofproblemsolving.com/9/2/c/92c82c921fad2971d842ec2dba87a5e4725c3e97.png)
Differentiability at 
implies 
is not possible.
and 
, reflecting about the line if necessary, we may assume 
.
and 
:![\[\usepackage{mathtools}
\tfrac {2\alpha+\eta-f(x)}{\alpha-x}\le 2\xRightarrow{\hspace{40pt}}\left(\forall 0<x<\alpha\right):f(x)\ \ge\ 2x+\eta\tag{3}\]](http://latex.artofproblemsolving.com/8/3/5/83584e1961403c7cf5c6891281a4cbb97346d193.png)
continuity at 
:![\[(\forall \epsilon>0)(\exists\delta>0):0<x<\delta\xRightarrow{\hspace{40pt}}\big|f(x)\big|<\epsilon\tag{4} \]](http://latex.artofproblemsolving.com/9/d/a/9da78481512b68b5c3e9ed771a83552d9878e137.png)
contradicts 
for 
.
is differentiable on . For 
and 
you can hopefully do!! put the value of 
in the two equations,So clearly visible that not differentiable on ....
and 
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and
is differentiable on . For and you can hopefully do!! put the value of in the two equations,So clearly visible that not differentiable on .... and 
be a function which is continuous on ![\( [0,1] \)](http://latex.artofproblemsolving.com/0/1/0/01026be294289be8f090c7b2586dde64a0d5b479.png)
and differentiable on 
,
.
such that 
.
such that 
