Problems For National: 2

[Masum]

Discuss the problems or post the solutions openly:

[sm.joty]

1.There should be little correction in Problem-4. I think it's a typo but better if Masum vai make us sure about this.
What about variable $a$ ?

2. Are you saying to establish a generalize formula for Problem-02 ???

[sm.joty]

I think this is the easiest Problem here.

Problem-03

The inequality equivalent to
$\frac{2}{\frac{1}{c}+\frac{1}{b}}+\frac{2}{\frac{1}{a}+\frac{1}{c}}+\frac{2}{\frac{1}{b}+\frac{1}{a}}\leq a+b+c$
Now,
$\frac{2}{\frac{1}{c}+\frac{1}{b}}\leq\frac{c+b}{2}...................(i)$
$\frac{2}{\frac{1}{a}+\frac{1}{c}}\leq\frac{a+c}{2}...................(ii)$
$\frac{2}{\frac{1}{b}+\frac{1}{a}}\leq\frac{b+a}{2}...................(iii)$

By adding all these inequality we get the desired result.

I'm not sure about it.
Problem-01:

We need to prove $gcd(a_n,a_{n-1})=1$ for all $n\geq3$

The base case $n=3$ is true.

Let
$gcd(a_n,a_{n-1})=1$ (induction hypothesis )
Then, $gcd(a_{n+1},a_n)=d$
As $a_{n+1}=3a_{n}+a_{n-1}$
and $d|3a_{n}$
so $d|a_{n-1}$
So $gcd(a_n,a_{n-1})=d$
According to induction hypothesis
$d=1$ and thus for all $n\geq3$, $gcd(a_n,a_{n-1})=1$
So, $gcd(a_{2011},a_{2012})=1$

[Masum]

I think this is the easiest Problem here.

Problem-03

The inequality equivalent to
$\frac{2}{\frac{1}{c}+\frac{1}{b}}+\frac{2}{\frac{1}{a}+\frac{1}{c}}+\frac{2}{\frac{1}{b}+\frac{1}{a}}\leq a+b+c$
Now,
$\frac{2}{\frac{1}{c}+\frac{1}{b}}\leq\frac{c+b}{2}...................(i)$
$\frac{2}{\frac{1}{a}+\frac{1}{c}}\leq\frac{a+c}{2}...................(ii)$
$\frac{2}{\frac{1}{b}+\frac{1}{a}}\leq\frac{b+a}{2}...................(iii)$

By adding all these inequality we get the desired result.

I'm not sure about it.
Problem-01:

and $d|3a_{n}$
so $d|a_{n-1}$

What about $d|3$? You can't conclude this.

[FahimFerdous]

I think problem 1 and 3 were easy. Problem 3 needs AM-GM. And as for problem 1, we can show that the gcd of any two consecutive terms divides all the terms. So, the gcd is actually 1. Still trying the others. And I think problem is really interesting and nice.

[sm.joty]

I think this is the easiest Problem here.

Problem-03

The inequality equivalent to
$\frac{2}{\frac{1}{c}+\frac{1}{b}}+\frac{2}{\frac{1}{a}+\frac{1}{c}}+\frac{2}{\frac{1}{b}+\frac{1}{a}}\leq a+b+c$
Now,
$\frac{2}{\frac{1}{c}+\frac{1}{b}}\leq\frac{c+b}{2}...................(i)$
$\frac{2}{\frac{1}{a}+\frac{1}{c}}\leq\frac{a+c}{2}...................(ii)$
$\frac{2}{\frac{1}{b}+\frac{1}{a}}\leq\frac{b+a}{2}...................(iii)$

By adding all these inequality we get the desired result.

I'm not sure about it.
Problem-01:

and $d|3a_{n}$
so $d|a_{n-1}$

What about $d|3$? You can't conclude this.

Masum vai, I said that $d|a_n$ thus $d|3a_n$
And $d|a_{n-1}=3a_n+a_{n-1}$
So $d|[3a_n+a_{n-1}-3a_n]$
so $d|a_{n-1}$

Hopefully there is no bug, if there then I think I can't understand that you say.

[Phlembac Adib Hasan]

মাসুম ভাই, এক ফোরামে পোস্ট কইরেন। খুঁজে পেতে সমস্যা হয়।

[Tahmid Hasan]

মাসুম ভাই,শেষ সমস্যায় $a,b$ এর প্রকৃতি কি?এরা যদি দুইটাই ধনাত্মক অথবা ঋণাত্বক পূর্ণসংখ্যা হয় তাহলে vieta jumping দিয়ে করা যায়।

[*Mahi*]

মাসুম ভাই,শেষ সমস্যায় $a,b$ এর প্রকৃতি কি?এরা যদি দুইটাই ধনাত্মক অথবা ঋণাত্বক পূর্ণসংখ্যা হয় তাহলে vieta jumping দিয়ে করা যায়।

That's something to be found out, not something to be told

[Nadim Ul Abrar]

SOL 4 :

$(a+b)^{2k}+a^{2k}+b^{2k} \equiv (a^2+ab+b^2+ab)^k+a^{2k}+b^{2k} \equiv a^{2k}+a^kb^k+b^{2k} mod (a^2+ab+b^2) $

Now $a^2+ab+b^2 \equiv 0 (mod p^x)$ ($x$ is the greatest integer such that for the prime $p$, $p^x|a^2+b^2+ab$)
then $(a-b)(a^2+ab+b^2) \equiv 0 (mod p^x)$
or $a^3 \equiv b^3(modp^x)$

putting$ k=3n+1$
$a^{2k}+a^kb^k+b^{2k} = a^{6n+2}+a^{3n+1}b^{3n+1}+b^{6n+2}$

$a^{6n+2}+a^{3n+1}b^{3n+1}+b^{6n+2}\equiv b^{6n}(a^2+ab+b^2) \equiv 0 (mod p^x)$ .

putting$ k=3n+2$
$a^{2k}+a^kb^k+b^{2k} = a^{6n+4}+a^{3n+2}b^{3n+2}+b^{6n+4}$

$a^{6n+4}+a^{3n+2}b^{3n+2}+b^{6n+4}\equiv b^{6n}(a^4+a^2b^2+b^4) \equiv b^2(ab+a^2+b^2)\equiv 0 (mod p^x)$ .

So we can say that for any prime $p$ if $p^x$ divide $a^2+ab+b^2$ then that will devide $(a+b)^{2k}+a^{2k}+b^{2k}$ too (if k is not divisible by 3),
$(a+b)^{2k}+a^{2k}+b^{2k} \geq a^2+ab+b^2$ So that $a^2+ab+b^2$ will divide $(a+b)^{2k}+a^{2k}+b^{2k}$

bug bug bug mammmmmiiiii