Discussion on Exam 1
Discussion on Exam 1
Ok, the first exam is over. You can discuss on the problems now. Solutions have been posted here. Until then, post your solutions and opinions on the problems or other solutions. And there will be no exam tomorrow. But there will one the next day after that. So be prepared for that.
One one thing is neutral in the universe, that is $0$.
Re: Discussion on Exam 1
Let me give you a hint on how to solve problem 1.2. Though I forgot to mention, but it should be kind of obvious that $a_0>1$.
Hint: If $s(a)$ is the smallest prime divisor of $a$, then $a+s(a)$ is always even.
Hint: If $s(a)$ is the smallest prime divisor of $a$, then $a+s(a)$ is always even.
One one thing is neutral in the universe, that is $0$.

 Posts: 1
 Joined: Mon Feb 18, 2013 5:57 pm
Re: Discussion on Exam 1
..........
Last edited by rubabredwan on Wed Aug 26, 2015 5:29 pm, edited 1 time in total.
Re: Discussion on Exam 1
Can anybody give me hints on how to solve problem 1.4?
Bored of being boring because being bored is boring
Re: Discussion on Exam 1
$a_0$ is a fixed positive integer which is given and greater than $1$. Try it again. And think about the hint too.rubabredwan wrote:Problem 1.2
$a_{0}$ can't be equal to 1.
let,
$a_{0}=2$ then,
$a_{1} = 4$
$a_{2} = 6/8$
$a_{3} = 8/9/10/12$
$a_{n} = ......$
if $a_{0}=3$ then,
$a_{1} = 6$
$a_{2} = 8/9$
$a_{3} = 10/12$
$a_{n} = ......$
we can see that integers 5, 7, 11 etc. aren't coming in this recurrence. because $a_{n}$ is a multiple of $d(a_{n1})$ which states that no prime other than $a_{0}$ will come.this recurrence follows a pattern like seive of erastothenes(not fully). if $a_{0}$ is not a prime, then every $a_{n}$ will be composite.
the number of prime numbers are infinite. So there is no such value of $k$.
One one thing is neutral in the universe, that is $0$.
Re: Discussion on Exam 1
$5$ divides $6^n1$.badass0 wrote:Can anybody give me hints on how to solve problem 1.4?
One one thing is neutral in the universe, that is $0$.
Re: Discussion on Exam 1
in problem 1.3 we just have to show that they are a primitive Pythagorean triple .
Re: Discussion on Exam 1
No, it is given that they are a Pythagorean Triple. You have to prove $2(b+p)$ is a perfect square.
One one thing is neutral in the universe, that is $0$.
Re: Discussion on Exam 1
i meant to say that if we can prove that they are primitive , then the work is almost done.Masum wrote:No, it is given that they are a Pythagorean Triple. You have to prove $2(b+p)$ is a perfect square.