and 
Prove that
in it. Prove that there are an infinite number of perfect cubes in this progression. (A perfect cube is an integer of the form 
,
is an integer. For example, 
,
and 
?
there are elements 
in an arithmetic progression in the order 
such that 
.
and 
be two sequences of positive numbers. Show that the following statements are equivalent:
of positive numbers such that 
and 
both converge;[/*]
converges.[/*]
be the number of positive divisors of 
dividing by x
be a positive root of the equation 
.
,
,
,
such that ![\[1 + xy + x^2y^2= a(x)c(y) + b(x)d(y)\]](http://latex.artofproblemsolving.com/4/d/7/4d74d8d44b9484443805d2b08e84c912aaa355a0.png)
holds identically?
![\[
\int_{1}^{2} \frac{x^{3}+x^2}{x^5}dx
\]](http://latex.artofproblemsolving.com/5/a/b/5ab9b8014b74dbfcf4db15873e0348bbdc0171ef.png)
![\[\int_{2025}^{2025^{2025}}\frac{1}{\ln\left(2025\right)\cdot x}dx\]](http://latex.artofproblemsolving.com/9/f/3/9f3fe22d62ad9fc5259d97fcf60eb3c45d240bfa.png)
![\[
\int(\sin^2(x)+\cos^2(x)+\sec^2(x)+\csc^2(x))dx
\]](http://latex.artofproblemsolving.com/8/3/d/83da36cded674b594ce7d585b1c1fee7df4956ae.png)
![\[
\int_{-2025.2025}^{2025.2025}\sin^{2025}(2025x)\cos^{2025}(2025x)dx
\]](http://latex.artofproblemsolving.com/9/1/0/91045d7c16160403507ee5f66a7cd021dfe0a0c3.png)
![\[
\int_{\frac \pi 6}^{\frac \pi 3} \tan(\theta)^2d\theta
\]](http://latex.artofproblemsolving.com/3/4/f/34f4936db8bf49e9d0c77de8539ba6ee144f94d5.png)
![\[
\int \frac{1+\sqrt{t}}{1+t}dt
\]](http://latex.artofproblemsolving.com/b/4/b/b4bead66ef90e5677281987993f82b1e4bb864df.png)
-----![\[\int \frac{x^{2}+2x+1}{x^{3}+3x^{2}+3x+3}dx
\]](http://latex.artofproblemsolving.com/9/2/2/922c2e35f18adffdd453ad926d0ae91fe759dd50.png)
![\[\int_{0}^{\frac{3\pi}{2}}\left(\frac{\pi}{2}-x\right)\sin\left(x\right)dx\]](http://latex.artofproblemsolving.com/3/6/b/36b02db24ba60cb50e0027a1d0d8a3ce1d07e72f.png)
![\[
\int_{-\pi/2}^{\pi/2}x^3e^{-x^2}\cos(x^2)\sin^2(x)dx
\]](http://latex.artofproblemsolving.com/a/c/5/ac50d60b5ff6d26d8ef2079efaaf7e2d293c8798.png)
![\[
\int\frac{1}{\sqrt{12-t^{2}+4t}}dt
\]](http://latex.artofproblemsolving.com/9/0/d/90d94cd0cdb89832dbf6b91dfb2af29169005963.png)
![\[
\int \frac{\sqrt{e^{8x}}}{e^{8x}-1}dx
\]](http://latex.artofproblemsolving.com/1/c/0/1c091415a17d162a745bd79dfb6c1632297d9ca0.png)
-----![\[
\int \frac{e^x}{e^{2x}+1} dx
\]](http://latex.artofproblemsolving.com/c/9/c/c9c93d4a2fc4e8856e0b6d30cffd5bf2581521bf.png)
![\[
\int_{-5}^5\sqrt{25-u^2}du
\]](http://latex.artofproblemsolving.com/7/6/0/760ca61e6e609ada7052eddb31c0b5cc8bdc7d8a.png)
![\[
\int_{-\frac12}^\frac121+x+x^2+x^3\ldots dx
\]](http://latex.artofproblemsolving.com/f/7/a/f7a1aa77b71a36c1800ed7bf5e44fa6744ee9a04.png)
![\[\int \cos(\cos(\cos(\ln \theta)))\sin(\cos(\ln \theta))\sin(\ln \theta)\frac{1}{\theta}d\theta\]](http://latex.artofproblemsolving.com/e/0/3/e030efdb1f1d73326dd09a60c961ac4b7bf987df.png)
![\[\int_{0}^{\frac{1}{6}}\frac{8^{2x}}{64^{2x}-8^{\left(2x+\frac{1}{3}\right)}+2}dx\]](http://latex.artofproblemsolving.com/4/a/2/4a274f4254113198782619e28ac63edcb41664ff.png)
-----![\[\int_{0}^{\frac{\pi}{2}}\frac{\sin\left(x\right)}{\sin\left(x\right)+\cos\left(x\right)}+\frac{\sin\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}dx\]](http://latex.artofproblemsolving.com/2/0/5/2051d51ec5ec892ce1f571385bfc1f74d05bea9f.png)
![\[
\int_0^{\infty}e^{-x}\Bigl(\cos(20x)+\sin(20x)\Bigr) dx
\]](http://latex.artofproblemsolving.com/4/e/4/4e482f5c117a0f3984c822614dd4a4214cc03b86.png)
![\[
\lim_{n\to \infty}\frac{1}{n}\int_{1}^{n}\sin(nt)^2dt
\]](http://latex.artofproblemsolving.com/3/b/3/3b3bf68b93dbe49e9979471ba04b82c3a2eb96c4.png)
![\[
\int_{x=0}^{x=1}\left( \int_{y=-x}^{y=x} \frac{y^2}{x^2+y^2}dy\right)dx
\]](http://latex.artofproblemsolving.com/c/0/3/c0314f295bc45fa87cfffc93ee348f38b44f51b2.png)
![\[
\int_{0}^{13}\left\lceil\log_{10}\left(2^{\lceil x\rceil }x\right)\right\rceil dx
\]](http://latex.artofproblemsolving.com/4/f/0/4f012045fcbc8210ea752729f609149b8a534e4f.png)
-----![\[
\int \frac{6x^2}{x^6+2x^3+2}dx
\]](http://latex.artofproblemsolving.com/0/d/1/0d1131a496881bcf7036de0b2a36191008925840.png)
![\[
\int -\sin(2\theta)\cos(\theta)d\theta
\]](http://latex.artofproblemsolving.com/b/0/0/b006ec3908ed9d9c00fe354578d50b5f91d56ddb.png)
![\[
\int_{0}^{5}\sin(\frac{\pi}2 \lfloor{x}\rfloor x) dx
\]](http://latex.artofproblemsolving.com/a/2/1/a21381f8fa2432165c02e7cd28dfb00ae3ad5475.png)
![\[
\int_{0}^{1} \frac{\sin^{-1}(\sqrt{x})^2}{\sqrt{x-x^2}}dx
\]](http://latex.artofproblemsolving.com/b/c/b/bcbe046b36c5d072bd6e064ee98a94683ccadfcc.png)
![\[
\int\left(\cot(\theta)+\tan(\theta)\right)^2\cot(2\theta)^{100}d\theta
\]](http://latex.artofproblemsolving.com/0/1/2/012796576e56002d2983da33436ddd6208ae3507.png)
-----![\[
\int \ln\left\{\sqrt[7]{x}^\frac1{\ln\left\{\sqrt[5]{x}^\frac1{\ln\left\{\sqrt[3]{x}^\frac1{\ln\left\{\sqrt{x}\right\}}\right\}}\right\}}\right\}dx
\]](http://latex.artofproblemsolving.com/d/6/f/d6fa1f1a12b988fef90f2a39f9b77f6e9822f76c.png)
![\[\int_{1}^{{e}^{\pi}} \cos(\ln(\sqrt{u}))du\]](http://latex.artofproblemsolving.com/e/1/5/e152dacb1601c772f32ab9f3494d45bed43c635f.png)
![\[
\int_e^{\infty}\frac {1-x\ln{x}}{xe^x}dx
\]](http://latex.artofproblemsolving.com/4/e/8/4e8ae310d63fae062122d6ff9ba1570ffaf98a42.png)
![\[\int_{0}^{1}\frac{e^{x}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{1}}}}\times\frac{e^{-\frac{x^{2}}{2}}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{2}}}}\times\frac{e^{\frac{x^{3}}{3}}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{3}}}}\times\frac{e^{-\frac{x^{4}}{4}}}{\left(x^{2}+3x+2\right)^{\frac{1}{2^{4}}}} \ldots \,dx\]](http://latex.artofproblemsolving.com/4/2/f/42feb97860007e1ef6ce36e03c71743b253f2576.png)
on the domain 
it is known that ![\[\displaystyle f(x)=\sin\left(\int_{0}^x \sqrt[3]{\cos\left(\frac{\pi}{2} t\right)^3+26}\ dt\right)\]](http://latex.artofproblemsolving.com/d/0/a/d0af1de2e29589d6205f05bad4cf0412742535b8.png)
is invertible. What is 
?
by![\[f(n) = \sum_{i\ge 0} \bigg\lfloor \frac{n + a^i}{a^{i+1}}\bigg\rfloor=\bigg\lfloor \frac{n + 1}{a} \bigg\rfloor + \bigg\lfloor \frac{n + a}{a^2} \bigg\rfloor + \bigg\lfloor \frac{n + a^2}{a^3} \bigg\rfloor + \cdots\]](http://latex.artofproblemsolving.com/8/f/5/8f5369fb9dfb998ea45bbe1ec16985cc6c2e8325.png)
for some fixed 
.![\[\lim_{n \rightarrow \infty} \frac{f(n)}{n/(a-1)} = 1.\]](http://latex.artofproblemsolving.com/0/6/c/06c0bf2d4619df935ccc5ad74832be520a796a74.png)
for 
.
be the set of all triples 
of positive integers for which there exist triangles with side lengths 
Express ![\[\sum_{(a,b,c)\in T}\frac{2^a}{3^b5^c}\]](http://latex.artofproblemsolving.com/9/f/6/9f6e0cee87e5b6958e143ff23b887bcd876f6cba.png)
as a rational number in lowest terms.
be a function such that 
and 
for all 
for some 
in domain then also we know second condition given in problem implies 
for all 
.
. Similarly, 
.
for all ![\[A=\begin{bmatrix} 1&1&1&1&1\\1&-1&1&-1&1\\?&?&?&?&?\\?&?&?&?&?\\?&?&?&?&?\end{bmatrix}\in\mathbb R^{5\times 5}\]](http://latex.artofproblemsolving.com/f/c/b/fcbdeb5107deefc1ec6441dfc0b7e93006729b9e.png)
satisfy 
Determine the maximum value of 
Y by
I am a little surprised to find that I am (so far) unable to solve this little problem:
I set
. If 
is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
Thank you.
Quote:
Let 
be an integer. Show that, if is an integer, then it is a perfect square.
be an integer. Show that, if 
is an integer, then it is a perfect square.I set
. If is an integer, then 
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
, and that, for every such pair 
, the "next" such pair can be calculated as
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?Thank you.
(because it is even), where 
is a positive integer.![\[
m - 1 = \sqrt{1+12n^2}
\]](http://latex.artofproblemsolving.com/1/6/3/16327bf406a6caceb2472cf6e97fab0f45289340.png)
![\[
(m-1)^2 = 1 + 12n^2
\]](http://latex.artofproblemsolving.com/b/e/7/be7dbc53a2157e3eaddf3a85aa71d139d35a29ca.png)
![\[
12n^2 = m(m-2).
\]](http://latex.artofproblemsolving.com/5/f/2/5f2e50576065c22b3863e838e9d365b79c74e8e5.png)
,
is a positive integer.![\[
12n^2 = 2k(2k-2)
\]](http://latex.artofproblemsolving.com/1/7/a/17a3e05e494322187c6e0c05c756a9f668a76e1d.png)
![\[
3n^2 = k(k-1).
\]](http://latex.artofproblemsolving.com/f/3/e/f3e013085fee695620922f5541d52c369ad17b2d.png)
and the left side is divisible by 3, exactly one of 
is divisible by 3.
.
,
is a positive integer.![\[2 + 2\sqrt{1+12n^2} = 2m = 4k = (2t)^2\]](http://latex.artofproblemsolving.com/e/7/5/e75c69c900bd846b088ee5910def581e49b74d50.png)
, which is a square.
is squarefree, and 
has primitive (meaning "smallest", based on, say, 
in positive integers, then the 
satisfies 
.
,
.
by 
and induct down)
which is fine, but calculating 
and 
