I have a confusion here. Can $x,y$ be negative?FahimFerdous wrote:Problem 38:
$3^x+7^y=n^2$
how many integer solutions for $(x,y)$ are there?
Secondary and Higher Secondary Marathon
Re: Secondary and Higher Secondary Marathon
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Re: Secondary and Higher Secondary Marathon
Firstly if $x,y$ are both nonnegative,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 $7^{2b+1}=n^23^{2a}=(n+3^{a})(n3^{a})$. Now we can assume that $n+3^{a}=7^{p},n3^{a}=7^{q},p+q=2b+1$. Subtraction implies that $2\times 3^{a}=7^{p}7^{q}$. If $q>0$ then $72\times 3^{a}$ 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?
$n^2\equiv 0(mod 4)$ 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 $7^{2b+1}=n^23^{2a}=(n+3^{a})(n3^{a})$. Now we can assume that $n+3^{a}=7^{p},n3^{a}=7^{q},p+q=2b+1$. Subtraction implies that $2\times 3^{a}=7^{p}7^{q}$. If $q>0$ then $72\times 3^{a}$ 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?
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with}} \color{green}{\textit{AVADA KEDAVRA!}}$

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Re: Secondary and Higher Secondary Marathon
I didn't understand how did you get that the $2$ perfect cube numbers.Tahmid Hasan wrote:There's a typo, it's from Iran NMO2008Tahmid 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 NMO20124.
Note: In BdMO Summer Camp2012, 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?Sorry, I don't but I remember Mahi and Nadim vai solved it using ring.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.
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 $a^6a^3+1$ is a perfect cube too.
$\forall a>1,a^31>0 \Rightarrow a^6a^3+1<(a^2)^3$
So $a^6a^3+1 \le (a^21)^3 \Rightarrow a^2(3a^2a3)+2 \le 0$.
But $\forall a>1, 3a^2>a+3$, so we have a contradiction.
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Re: Secondary and Higher Secondary Marathon
See the statement at first:sakib.creza wrote: I didn't understand how did you get that the $2$ perfect cube numbers.
Substitute $n=3,9$.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 NMO20124.
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 Phlembac Adib Hasan
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Re: Secondary and Higher Secondary Marathon
$n$এর সংজ্ঞা কি একই রাখা হবে না পাল্টানো হবে? মানে পূর্ণসংখ্যার বদলে মূলদ বলা হইলে কষ্ট বেশি হবে।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?
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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Re: Secondary and Higher Secondary Marathon
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
 samiul_samin
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Re: Secondary and Higher Secondary Marathon
Rewriting the problem:samiul karim wrote: ↑Mon Dec 21, 2015 4:27 pmProve 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
$H_n=\dfrac 11 +\dfrac 12+\dfrac 13 +...+\dfrac 1n$
Prove that,if $n>1$,then $H_n$ is not equal to any integer.
Problem source