Secondary and Higher Secondary Marathon

For students of class 11-12 (age 16+)
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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Tue Feb 05, 2013 11:16 pm

FahimFerdous wrote:Problem 38:

$3^x+7^y=n^2$
how many integer solutions for $(x,y)$ are there?
I have a confusion here. Can $x,y$ be negative? :? :?
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SANZEED
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Re: Secondary and Higher Secondary Marathon

Unread post by SANZEED » Tue Feb 05, 2013 11:54 pm

Firstly if $x,y$ are both non-negative,then we can do the following:
$n^2\equiv 0(mod 4)$ here.And also $3\equiv -1(mod 4)$ and $7\equiv -1(mod 4)$. So one of $x,y$ is odd and the other is even. Let $x=2a,y=2b+1$. Then $7^{2b+1}=n^2-3^{2a}=(n+3^{a})(n-3^{a})$. Now we can assume that $n+3^{a}=7^{p},n-3^{a}=7^{q},p+q=2b+1$. Subtraction implies that $2\times 3^{a}=7^{p}-7^{q}$. If $q>0$ then $7|2\times 3^{a}$ which is not possible. So $q=0,p=2b+1$. Thus we get $2\times 3^{a}+1=7^{2b+1}=(2\times 3+1)^{2b+1}$. Expanding the RHS shows that $a=1$ and consequently $b=0$. Thus $(2,1)$ is the only solution. Similarly we can assume that $x=2a+1,y=2b$ and a similar argument implies that $(1,0)$ is another solution.

However,It is not clear to me whether $x,y$ can be negative or not. So I can't post anything for negative $x,y$. Can anyone help me please? :? :?
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sakib.creza
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Re: Secondary and Higher Secondary Marathon

Unread post by sakib.creza » Wed Feb 06, 2013 7:09 am

Tahmid Hasan wrote:
Tahmid Hasan wrote:Probelm $37$: $\mathbb{N}$ is the set of positive integers and $a\in\mathbb{N}$. We know that for every $n\in\mathbb{N}$, $4(a^n+1)$ is a perfect cube. Prove that $a=1$.
Source: Iran NMO-2012-4.
Note: In BdMO Summer Camp-2012, a similar problem was given in the Number Theory problem set.
Zubaer vai gave a 'cruxy' solution, but unfortunately I don't remember it. So I solved it with brute force. :(
Does anybody remember?
There's a typo, it's from Iran NMO-2008 :oops:
Phelembac Adib Hasan wrote:@Tahmid vai, please post your proof and if you can still remember Zubayer vai's nice proof, please post it, too.
Sorry, I don't :( but I remember Mahi and Nadim vai solved it using ring.
My solution: Note that if $a=1,4(a^n+1)=8=2^3$.
Now assume $a>1$.
$4(a^9+1),4(a^3+1)$ are both perfect cubes, so their quotient $a^6-a^3+1$ is a perfect cube too.
$\forall a>1,a^3-1>0 \Rightarrow a^6-a^3+1<(a^2)^3$
So $a^6-a^3+1 \le (a^2-1)^3 \Rightarrow a^2(3a^2-a-3)+2 \le 0$.
But $\forall a>1, 3a^2>a+3$, so we have a contradiction.
I didn't understand how did you get that the $2$ perfect cube numbers.

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Phlembac Adib Hasan
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Re: Secondary and Higher Secondary Marathon

Unread post by Phlembac Adib Hasan » Wed Feb 06, 2013 9:45 am

sakib.creza wrote: I didn't understand how did you get that the $2$ perfect cube numbers.
See the statement at first:
Tahmid Hasan wrote:Probelm $37$: $\mathbb{N}$ is the set of positive integers and $a\in\mathbb{N}$. We know that for every $n\in\mathbb{N}$, $4(a^n+1)$ is a perfect cube. Prove that $a=1$.
Source: Iran NMO-2012-4.
Substitute $n=3,9$.
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Re: Secondary and Higher Secondary Marathon

Unread post by Phlembac Adib Hasan » Wed Feb 06, 2013 10:00 am

SANZEED wrote:However,It is not clear to me whether $x,y$ can be negative or not. So I can't post anything for negative $x,y$. Can anyone help me please? :? :?
$n$-এর সংজ্ঞা কি একই রাখা হবে না পাল্টানো হবে? মানে পূর্ণসংখ্যার বদলে মূলদ বলা হইলে কষ্ট বেশি হবে।
Consider these cases:
$1.\; x=0, z=-y>0\, \, \Longleftrightarrow 7^{z}+1=n^27^z$
আরও তিন চারটা কেস আসবে। সমাধানের বিস্তৃতি হবে ইনশাল্লাহ আমার জুনিয়র বলকানের সমান। আমার এখন করার কোন ইচ্ছা নাই। আর কেউ চাইলে করতে পারেন।
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samiul karim
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Re: Secondary and Higher Secondary Marathon

Unread post by samiul karim » Mon Dec 21, 2015 4:27 pm

Prove that 1+1/1+1/2+1/3+1/4+1/5+..........1/n not equal any integer i have proved that but not sure about the way of proving

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Re: Secondary and Higher Secondary Marathon

Unread post by samiul_samin » Fri Feb 23, 2018 9:01 am

samiul karim wrote:
Mon Dec 21, 2015 4:27 pm
Prove that 1+1/1+1/2+1/3+1/4+1/5+..........1/n not equal any integer i have proved that but not sure about the way of proving
Rewriting the problem:

$H_n=\dfrac 11 +\dfrac 12+\dfrac 13 +...+\dfrac 1n$

Prove that,if $n>1$,then $H_n$ is not equal to any integer.

Problem source
Olympiad Somogro
Solution
Let,$H_n=\dfrac xy$
If $n>1$ then $y$ is always even.
But $x$ is always odd.
Because,multiple of some primes without $2$ is always odd.
So,$\dfrac xy$ can never be an integer.Because,no odd number is divided by any even number.
So,$H_n$ is not an integer if $n>1$[Q.E.D]

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