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?
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
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
Last edited by Mehrab4226 on Sun Mar 21, 2021 8:34 pm, edited 1 time in total.
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. -Henri Poincaré
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
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. -Henri Poincaré
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 $\tau(n)<2\sqrt{n}$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
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 $\tau(n)<2\sqrt{n}$
Ahh!!. yes, yes. Clever one!
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful. -Henri Poincaré
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.$\blacksquare$
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 $p≠2,13$
which doesn't divide $5^2+1$
And we are done.
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.$\square$
Last edited by Asif Hossain on Wed Mar 24, 2021 9:33 pm, edited 7 times in total.