BdMO 2010 H. Sec. problem 8

[sm.joty]

I am posting to find all such primes $p$. Let wlog $a\ge b$ (even if both negative or one positivd another negative). If $a=b,p=2$.

মাসুম ভাই,
১.এখানে, $a=b$ হলে $p=2$ কেন বুঝলাম না।

Now let $a=b+c,c>0$.So $p^b(p^c+1)$ is a rational square.Since $gcd(p^b,p^c+1)=1$ when $b>0$or $gcd(p^d,p^c+1)=1$ when $b=-d,d>0$; we see that $p^c+1$ is a square.

২.এখানে কি ধরে নিয়েছেন যে $a=b+c,c>0$ এর জন্য $p^b(p^c+1)$ একটা মূলদ বর্গ ?
৩."$gcd(p^d,p^c+1)=1$ when $b=-d,d>0$; we see that $p^c+1$ is a square."
এই লাইনের কিছুই বুঝি নাই ?

Then $k+1=p^x,k-1=p^y;p^x-p^y=p^y(p^{x-y}-1)=2 \Longrightarrow p=2,y=1$ or, $y=0,p=3,x=1$.

৪.$p^x-p^y=p^y(p^{x-y}-1)=2$ কেন হল, মানে ২ কেন আসল ?
৫.$p=2,y=1$ or, $y=0,p=3,x=1$ ঐ সম্পর্ক থেকে কিভাবে এই সমাধান গুলা পেলেন। trail & error method নাকি অন্য কোন পদ্ধতিতে ?

বি. দ্র ঃ আমার মাথায় কিচ্ছু নাই। তাই এসব প্রশ্ন করলাম ?

[Abdul Muntakim Rafi]

Blah blah blah!!! not related to solution... So hid this part...

Sorry... I can't use equation editor for writing all these equations... However,as I have used logic in spite of equation that shouldn't be a problem... I solved this problem a year ago... After a lot of discussion(in yahoo math club and groups)Posting that solution...

The solution is for not rational numbers but for natural numbers(as Nayel bhaiya said in the hint)... That problem was for BDMO 2010 Secondary 8 or 9 or 10th problem...

Let's consider 2 cases.a=b & a is not equal to b.(Actually Moon bhai gave me this start) consider two cases...

Let a=b,
Then $2p^a=n^2 $
Let's observe it.If we put p=3,5,7....odd
primes we won't be able to find any square number.But if we use
p=2,a=3,,5,7,9.....we will get a square number.That's the key.We
multiply the $2$ with $2,8,32$......I mean $2^1,2^3,2^5$.... That will give us $2^2,2^4,2^6$ which are squares...
so here p=2 and
a=positive odd.

So we get an answer.That p=2 and a,b=positive odd.a,b are the same odd number at one time

Now for a not equals to b... Now for our benefit we assume that a is less than b(b could be less than a,that does not make any difference as far as the above question concerned) Here we need to be tricky.
$p^a+p^a*p^b-a=p^a(1+p^b-a)=p^a(1+p^c) [let,b-a=c]$
As it equals to a square and
they both are relative primes,so they have to be square numbers themselves. That
means $p^a$ and $1+p^c$ are squares.Let
$1+p^c=x^2$
or,$p^c=x^2-1=(x+1)(x-1)$
Now take 2 cases.p odd and p even.
if p is odd.p=3.The reason is that
$p^c=(x+1)(x-1)$
As p^c is odd these both are odd too.their
gcd=1.
(For determining the smallest possible values)
And one of them can be 1.Clearly x-1=1 so,x=2 & x+1=3 that implies
p=3.c=1 now c=b-a=1 so,b=a+1 ....
we can see that a=even and b=the odd
next to a,I mean consequent number......
so p=3,a=2k and b=2k+1
Bingo!

Like it prove that for p even

In this case $2^c=(x+1)(x-1)....here gcd(x+1,x-1)=2$....
as 2^c is even these both are even too...one of them is even for sure... that makes the other an even number...as their difference is 2...that's why here gcd=2.

now considering lowest values
$x-1=2,x=3,b-a=3,b=a+3$ so here p=2,a=2k,b=2k+3 here k belongs to N as a,b belongs to N.

Bingo!We solved it.

[Masum]

I am posting to find all such primes $p$. Let wlog $a\ge b$ (even if both negative or one positivd another negative). If $a=b,p=2$.

মাসুম ভাই,
১.এখানে, $a=b$ হলে $p=2$ কেন বুঝলাম না।
If $a=b$, then we need $p^a+p^b=2p^a$ an even perfect square, so divisible by $4$, so $p$ even and hence $p=2$.

Now let $a=b+c,c>0$.So $p^b(p^c+1)$ is a rational square.Since $gcd(p^b,p^c+1)=1$ when $b>0$or $gcd(p^d,p^c+1)=1$ when $b=-d,d>0$; we see that $p^c+1$ is a square.

২.এখানে কি ধরে নিয়েছেন যে $a=b+c,c>0$ এর জন্য $p^b(p^c+1)$ একটা মূলদ বর্গ ?
Note that a positive integer $a$ and $p$, a prime has the greatest common divisor either $p$ or $1$. In this case $1$ since $p^c+1$ yields a remainder $1$ upon division by $p$. So, every $p$ of $p^b=p....p$ is co-prime to $p^c+1$. Then $p^b$ is co-prime to $p^c+1$. If $b$ is positive, then $p^b$ and $p^c+1$ both are perfect squares(Because if $xy=a^2$ with $x$ and $y$ sharing no common factor then $x$ and $y$ both are squares as well). Then we need $b$ even, but that is not that much headache. The ache is to find $c$. Got it?
৩."$gcd(p^d,p^c+1)=1$ when $b=-d,d>0$; we see that $p^c+1$ is a square."
এই লাইনের কিছুই বুঝি নাই ?
If it is a rational square, then we would have $b$ negative, say $b=-d$ where $d$ is positive. So \[\frac{p^c+1}{p^d}\] is a perfect square. Again similarly, $p^c+1$ and $p^d$ co-prime. Then again we need to find $c$ only.

Then $k+1=p^x,k-1=p^y;p^x-p^y=p^y(p^{x-y}-1)=2 \Longrightarrow p=2,y=1$ or, $y=0,p=3,x=1$.

৪.$p^x-p^y=p^y(p^{x-y}-1)=2$ কেন হল, মানে ২ কেন আসল ?
Subtract the second equation from the first one.
৫.$p=2,y=1$ or, $y=0,p=3,x=1$ ঐ সম্পর্ক থেকে কিভাবে এই সমাধান গুলা পেলেন। trail & error method নাকি অন্য কোন পদ্ধতিতে ?
If $y>0$ then $p|2$ which gives a contradiction. So $y=0$ Clear?