NUMBER THEORY MARATHON: SEASON 2

[sourav das]

I guess it's time to restart Number Theory Marathon.

RULES: Very simple. I'll start this Marathon with a number theory problem. Next who can give the solution of the problem, will get a chance to post a new problem. Different solutions are welcome too. Who will give a different solution also can get a chance to post a new problem. To make things interesting, one have to post solution of a problem within 12 hours. After 12 hours, the problem will be a star problem and the problem will be re-posted in a new topic. But the Number Theory Marathon will be continued by posting a new problem from the former solvers....

Hope u will enjoy the Number Theory Marathon.

Problem 1:
Determine all primes $p$ such that $5^p+ 4p^4$ is a square number.

[*Mahi*]

Let $5^p+4p^4=k^2$.
Then,$5^p=k^2-4p^4$or $5^p=(k+2p^2)(k-2p^2)$
As $5$ is prime, $k+2p^2=5^i$ and $k-2p^2=5^{p-i}$ where $p \geq 2i$
So $4p^2=5^{p-i}-5^i$
or, $4p^2=5^i(5^{p-2i}-1)$
Or,$p^2=5^i( \sum_{j=1} ^{p-2i} 5^{p-2i-j} )$
So,$5|p$ or $5=p$.

[*Mahi*]

There is another case $i=0$.
So, $5^p=4p^2+1$
Again $5^p \equiv 5 (mod$ $p)$
So, $4p^2+1 \equiv 1 \equiv 5$ $(mod$ $p)$ which can't be true.

[sourav das]

Post your problem please

[*Mahi*]

New Problem:
Prove that, the equation $x^4-y^4=z^2$ is not solvable in nonzero integers.

[Tahmid Hasan]

New Problem:
Prove that, the equation $x^4-y^4=z^2$ is not solvable in nonzero integers.

we'll us Euler's infinite descence:first think the solution are symmetric in Y-axis.
so we'll consider positive integers.
i'll use Infinite descence:let us assume $(x,y,z)$ are solutions for which $x$ is the smallest.now we'll show another $x$ smaller than the previous contradicting the statement.
let $gcd(x,y)=m$ then $m^4 \mid x^4,m^4 \mid y^4,m^4 \mid z^2$,
or,$\frac{x^4}{m4}- \frac{y^4}{m^4}=\frac{z^2}{m^4}$
so a contradiction.
so $x,y$ are co-prime.we e-write the equation as $y^4+z^2=x^4$,hence $(y^2,z,x^2)$ are primitive pythagorean triples.
if $y$ is odd,then there exists natural $p,q$ such that $y^2=p^2-q^2,x^2=p^2+q^2$
then $p^4-q^4=(xy)^2$,since $p<x$,this is a contradiction
if $y$ is even, there exist co-prime $p,q$ such that $x^2=p^2+q^2,y^2=2pq$,by symmetry let's assume $p,q$ are odd and even respectively.
so $(p,q,x)$ are primitive pythagorean triples.hence there are co-prime $a,b$ such that $p^2=a^2-b^2,q=2ab$
but $y^2=2pq=2(a^2-b^2).2ab$,so $p=c^2.q=2d^2$,since $q=2ab=2d^2$,then $a,b$ are both perfect squares.
let $a=u^2,b=v^2$,then $p=c^2=a^2-b^2=u^4-v^4$,which is a contradiction.

[Tahmid Hasan]

new problem:
find all integers $(a,b)$ such that $\frac{a^7-1}{a-1}=b^5-1$

[Masum]

Lemma: Every prime divisor $p$ of $\frac{x^q-1}{x-1}$ is either $p=q$ or $p\equiv1\pmod q$.(Try to prove)
This lemma kills the problem. So, I am leaving the rest too.
I have no problem to give.

BTW, this is a problem from IMO-Shortlist, 2006.

[Masum]

New Problem:
Prove that, the equation $x^4-y^4=z^2$ is not solvable in nonzero integers.

This is a special case of Fermat's Last Theorem( when you set $z=t^2$). And the case for $n$ is denoted by $\text{FLT}_n$. So using this lemma, $\text{FLT}_4$ is proved too. However, you may see this for further reading:

[Masum]

It seems that I have almost killed the marathon.
T continue this, I have a problem. Recently, I found the following theorem:
The number of numbers having an even sum of divisors less than or equal to $n$ is $n- \lfloor\sqrt n\rfloor-\lfloor\sqrt{\frac{n}{2}}\rfloor$.
You can find a hint here
In fact this is the key lemma needed to prove the theorem.
Now prove it.