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?
If this quadratic equation has solutions in integers, then its discriminant must be a perfect square.
So, $b^2-4ac=6^2-4*12*1=36-48$ is not a perfect square.
Isn't this the case when we are trying to solve $12b^2+6b+1=0$? But in this case,we are trying to find $b$ such that $12b^2+6b+1$ is a perfect square, not the root of that polynomial, so why is discriminant so important?
Solution of Problem:2
Posted: Sun Mar 21, 2021 8:19 pm
by Mehrab4226
Let,
$d_1,d_2,d_3 \cdots d_k$ be the divisiors of $n$ not greater than $\sqrt{n}$.
$\therefore \frac{n}{d_1},\frac{n}{d_2},\cdots \frac{n}{d_k}$ are the other divisors.
$\therefore \tau (n) \leq 2k$ [Less is when n is a square number]
But, $k \leq \sqrt{n}$ because by definition k cannot exceed $\sqrt{n}$
$\therefore \tau (n) \leq 2\sqrt{n}$
I probably read the solution before in a book. Probably
Problem:3
Posted: Sun Mar 21, 2021 8:25 pm
by Mehrab4226
Let $n$ be a positive integer and let $a_1,a_2,a_3,\cdots ,a_k$($k\geq 2)$ be distinct integers in the set $1,2,\cdots , n$ such that $n$ divides $a_i(a_{i+1}-1)$ for $i=1,2,\cdots , k-1$. Prove that $n$ does not divide $a_k(a_1-1).$
Let,
$d_1,d_2,d_3 \cdots d_k$ be the divisiors of $n$ not greater than $\sqrt{n}$.
$\therefore \frac{n}{d_1},\frac{n}{d_2},\cdots \frac{n}{d_k}$ are the other divisors.
$\therefore \tau (n) \leq 2k$ [Less is when n is a square number]
But, $k \leq \sqrt{n}$ because by defination k cannot exceed $\sqrt{n}$
$\therefore \tau(n) \leq 2\sqrt{n}$
I probably read the solution before in a book. Probably
From here, we can easily see that the equality never holds. Cause if $n$ is a perfect square, then $\tau(n)$ is odd and $2\sqrt{n}$ is even. So it must always be the case that $\tau(n)<2\sqrt{n}$
Let,
$d_1,d_2,d_3 \cdots d_k$ be the divisiors of $n$ not greater than $\sqrt{n}$.
$\therefore \frac{n}{d_1},\frac{n}{d_2},\cdots \frac{n}{d_k}$ are the other divisors.
$\therefore \tau (n) \leq 2k$ [Less is when n is a square number]
But, $k \leq \sqrt{n}$ because by defination k cannot exceed $\sqrt{n}$
$\therefore \tau(n) \leq 2\sqrt{n}$
I probably read the solution before in a book. Probably
From here, we can easily see that the equality never holds. Cause if $n$ is a perfect square, then $\tau(n)$ is odd and $2\sqrt{n}$ is even. So it must always be the case that $\tau(n)<2\sqrt{n}$
Let $n$ be a positive integer and let $a_1,a_2,a_3,\cdots ,a_k$($k\geq 2)$ be distinct integers in the set $1,2,\cdots , n$ such that $n$ divides $a_i(a_{i+1}-1)$ for $i=1,2,\cdots , k-1$. Prove that $n$ does not divide $a_k(a_1-1).$
Source:
IMO SL 2009, N1
$\textbf{Solution 3}$
We have, $ a_{i-1}(a_i-1)\equiv 0\pmod{n}\Rightarrow a_ia_{i-1}\equiv a_{i-1}\pmod{n}$
So, $a_ka_{k-1}a_{k-2}\cdots a_1\equiv a_{k-1}a_{k-2}\cdots a_1\equiv a_{k-2}a_{k-3}\cdots a_1\equiv\cdots \equiv a_1 \pmod{n}$
Going on like this at point we'll have $a_ka_{k-1}a_{k-2}\cdots a_2 a_1\equiv a_2 \pmod{n}$
Therefore, $a_1\equiv a_2\pmod{n}$, but this cannot be possible as both of them are smaller than $n$, so they cannot have same remainder upon dividing by $n$, Contradiction.$\blacksquare$
Re: NT marathon!!!!!!!
Posted: Sun Mar 21, 2021 10:48 pm
by ~Aurn0b~
$\textbf{Problem 4}$
Solve the equation
\[ 3^x - 5^y = z^2.\]
in positive integers.
Re: NT marathon!!!!!!!
Posted: Mon Mar 22, 2021 1:14 pm
by Dustan
Solution 4: taking mod 4
$(-1)^x-1\cong 0/1(mod4)$
In R.H.S. $z^2$ should be $\cong$to 0. Otherwise
$(-1)^x=1+1\cong 2$ contradiction.Since L.H.S. can be 3 or 1 only.
So,$x$ is even. Hence $z$ also.
Let, $x=2k$
$(3^k+z)(3^k-z)=5^y$,where
$(3^k+z,3^k-z)=(3^k-z,2z)=(3^k,z)=1$
So, $(3^k-z)=1 $ and $(3^k+z)=5^a $ for some a
Now, $a=1$ gives $(x,y,z)=(2,1,2)$
$a=2$ has no integer soln.
For $3\leq a$ according to the zsigmondy theorem
There will exist atleast a prime factor $pā 2,13$
which doesn't divide $5^2+1$
And we are done.
(Is there any mistake?):(
Re: NT marathon!!!!!!!
Posted: Tue Mar 23, 2021 8:33 pm
by Asif Hossain
Sorry for my wrong solu of prob 1 here is my another elementary solu:(pls confirm)
Taking mod $4$ it is clear that $x$ must be odd. Case 1: $y$ is odd.
We can rewrite the equation as $(x-1)(x^2+x+1)=2y^2$ now since $x^2+x+1$ is odd then $(x^2+x+1)|y^2$ now by little modular arithmetic it is easy to see $x \equiv 1 (mod 4)$ or $(x-1) \equiv 0 (mod4)$ but $(x-1) \equiv 0 (mod2)$ which implies $y^2$ is even which contradicts that $y$ is odd, Case2: $y$ is even.
.Let $x=2k_1 +1$ and $y=2k_2$
Plugging it into the original equation
$\Rightarrow 8k_{1}^{3}+12k_{1}^{2}+6k_1=8k_{2}^{2}$
Now taking mod 4 both side implies $6k_1 \equiv 0 \Rightarrow 2k_1 \equiv 0 (mod 4)$ which implies $k_1$ is even since all positive integer
Then also taking modulo 8 implies $k_1(12k_1+6) \equiv 0 (mod 8)$ one case is $12k_1+6 \equiv 0 (mod 8)$ which implies $k_1$ is odd so contradiction
the other case is $k_1 \equiv 0 (mod 8)$ or $(mod4)$ WLOG that is not possible since iterating the process avoiding odd value of $k_1$ would then mean $k_1$ is divisible by any power of $2$ so then it would have no solution.$\square$
