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NT marathon!!!!!!!
Posted: Fri Mar 19, 2021 3:07 pm
by Dustan
HI , NT MARATHON is here. Rules are same as the previous marathons. Happy Problem Solving!
$Problem 1 $: Find all positive integer x,y such that
$x^3=2y^2+1$
Re: NT marathon!!!!!!!
Posted: Sat Mar 20, 2021 7:43 pm
by Dustan
$x$ is odd.
Now,
$2y^2+1$=$(1+y\sqrt{-2})(1-y\sqrt{-2}$
this two are coprime elements of $\mathbb{Z}[\sqrt{-2}]$
so ,both of them are perfect cube
So, there exist a,b in integer such that
$(a+b\sqrt{-2})=(1+y\sqrt{-2})$
Comparing the coefficients at 1 and $\sqrt{-2}$ yields
$a^3-6ab^2=1$
hence $(a,b)=(1,0)$
$x=1$ and $y=0$ but $y$ isn't a positive integer.
THERE IS NO SOLUTION.
Re: NT marathon!!!!!!!
Posted: Sat Mar 20, 2021 8:11 pm
by Mehrab4226
Dustan wrote: ↑Sat Mar 20, 2021 7:43 pm
$x$ is odd.
Now,
$2y^2+1$=$(1+y\sqrt{-2})(1-y\sqrt{-2}$
this two are coprime elements of $\mathbb{Z}[\sqrt{-2}]$
Could you explain what the notation $\mathbb{Z}[\sqrt{-2}]$ means?
So, there exist a,b in integer such that
$(a+b\sqrt{-2})=(1+y\sqrt{-2})$
And how did you get that?
Re: NT marathon!!!!!!!
Posted: Sat Mar 20, 2021 8:55 pm
by Asif Hossain
Mehrab4226 wrote: ↑Sat Mar 20, 2021 8:11 pm
Dustan wrote: ↑Sat Mar 20, 2021 7:43 pm
$x$ is odd.
Now,
$2y^2+1$=$(1+y\sqrt{-2})(1-y\sqrt{-2}$
this two are coprime elements of $\mathbb{Z}[\sqrt{-2}]$
Could you explain what the notation $\mathbb{Z}[\sqrt{-2}]$ means?
So, there exist a,b in integer such that
$(a+b\sqrt{-2})=(1+y\sqrt{-2})$
And how did you get that?
Maybe he is referring
about gaussian integers.
Problem 2
Posted: Sat Mar 20, 2021 9:41 pm
by Anindya Biswas
Let $\tau:\mathbb{N}\to\mathbb{N}$ such that $\tau(n)$ is the number of positive divisors of $n$. Prove that $\tau(n)<2\sqrt{n}$.
Re: NT marathon!!!!!!!
Posted: Sat Mar 20, 2021 10:35 pm
by Asif Hossain
This can maybe a solution of prob 1(I don't know really NT noob)(Kindly correct me if i am wrong)
Re: NT marathon!!!!!!!
Posted: Sat Mar 20, 2021 11:04 pm
by Dustan
One of my friend gave me cool solution
.
Here is the soln
$2y^2+1=8a^3+12a^2+6a+1$
Or,$Y^2=a(4a^2+6a+3)$
$(a, 4a^2+6a+3)=1\ or 3$
Case1: , gcd is 1
Both of them are perfect square.
But $(2a+1)^2<4a^2+6a+3<(2a+2)^2$
Contradiction.
Case 2: gcd is 3
$a=3b$
$3b(36a^2+18a+3)=9b(12b^2+6b+1)$
As,$(b, 12b^2+6b+1)=1$
Both of them are square.
From the discriminate, we can say $12b^2+6b+1$ is not a square.
Hence we are done
Re: NT marathon!!!!!!!
Posted: Sat Mar 20, 2021 11:49 pm
by Anindya Biswas
Asif Hossain wrote: ↑Sat Mar 20, 2021 10:35 pm
Then also taking modulo 8 implies $k_1(12k_1+6) \equiv 0 \Rightarrow 12k_1+6 \equiv 0 (mod 8)$
I don't think $k_1$ can be cancelled like that. Cause in order to cancel it, you must need $\text{gcd}(k_1,8)=1$ which is not the case since you proved $k_1$ is even.
Re: NT marathon!!!!!!!
Posted: Sun Mar 21, 2021 12:05 am
by Anindya Biswas
Dustan wrote: ↑Sat Mar 20, 2021 11:04 pm
From the discriminate, we can say $12b^2+6b+1$ is not a square.
I think this sentence needs some clearance... How can we say that $12b^2+6b+1$ is never going to be a perfect square?
Re: NT marathon!!!!!!!
Posted: Sun Mar 21, 2021 7:54 am
by Asif Hossain
Anindya Biswas wrote: ↑Sat Mar 20, 2021 11:49 pm
Asif Hossain wrote: ↑Sat Mar 20, 2021 10:35 pm
Then also taking modulo 8 implies $k_1(12k_1+6) \equiv 0 \Rightarrow 12k_1+6 \equiv 0 (mod 8)$
I don't think $k_1$ can be cancelled like that. Cause in order to cancel it, you must need $\text{gcd}(k_1,8)=1$ which is not the case since you proved $k_1$ is even.
Sorry it need some clarification
now
I also mentioned the case $k \equiv 0 (mod 8)$ or $(mod4)$ afterwards
.