Find all non negative integer (a,b) such that.....

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AourkoPChakraborty
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Joined:Tue Feb 09, 2021 5:23 pm
Find all non negative integer (a,b) such that.....

Unread post by AourkoPChakraborty » Mon Feb 15, 2021 12:38 am

Find all non negative integers $(a,b)$ such that, $3^a +7^b$ is a perfect square.

Source: Canada Mathematical Olympiad 2009, p4.

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Find all non negative integer (a,b) such that.....

Unread post by Asif Hossain » Sat Mar 13, 2021 11:11 pm

THERE IS NO PROOF
Seriously
THERE is no proof
believe me
:|
OK it's your choice
Hmm..Hammer...Treat everything as nail

Asif Hossain
Posts:194
Joined:Sat Jan 02, 2021 9:28 pm

Re: Find all non negative integer (a,b) such that.....

Unread post by Asif Hossain » Thu Apr 15, 2021 10:37 am

AourkoPChakraborty wrote:
Mon Feb 15, 2021 12:38 am
Find all non negative integers $(a,b)$ such that, $3^a +7^b$ is a perfect square.

Source: Canada Mathematical Olympiad 2009, p4.
Actual proof:
Let $3^a +7^b =x^2$
Taking modulo $4$ it is clear that $a,b$ are of different parity and $x$ is even.
Case 1: $(a,b)=(a,0)$
Then the expression yields $3^a =(x+1)(x-1)$ since $3$ doesn't divide both $x+1$ and $x-1$ the only possible case is either $x+1=1 \Rightarrow x=0$ which doesn't work
Or, $x-1=1 \Rightarrow x=2 \Rightarrow a=1$
So, one of solution is $(1,0)$
Case 2: $(a,b)=(0,b)$
Then the expression yields $7^b=(x+1)(x-1)$ since $3$ doesn't divide both $x+1$ and $x-1$ the only possible case is either $x+1=1 \Rightarrow x=0$ which doesn't work
Or, $x-1=1 \Rightarrow x=2$ which also doesn't work.
So, there is no solution for this case.

Now, we will find solution for positive integer.
Case 3:$b$ is even and $a$ is odd.
Let $b=2y$ and $a=2k+1$
The expression yields $3^a = (x+7^y)(x-7^y)$ since the $3$ can't divide both $x+7^y$ and $x-7^y$ the only possiblity is either $x+7^y=1$ which is not true for positive $y$.
OR,$x-7^y=1$ and $x+7^y=3^a$. This yields $x \equiv1 (mod 7)$
So, $3^a \equiv1 (mod 7)$ which is not possible for odd $a$
So, no solution for this case.

Case 4:$a$ is even and $b$ is odd.
Now, let $a=2y$ and $b=2k+1$
Again by the same arguement of Case 3 we can only get $x-3^y=1$ and $x+3^y=7^b$
Combining them we get $2.3^y=7^b -1$
By LTE, we can get $\nu_{3}(b)=y-1$
FTSOC, assume $3|b$ so $b=3k$ where $k$ is odd and $y>1$
we get $x \equiv 1 (mod 9)$ from the first equation and $x \equiv -1 (mod 9)$ from the second.Contradiction.
So, $y-1=0 \Rightarrow y=1$
So,$(a,b)=(1,0),(2,1)$ are the only solutions. $\square$
Hmm..Hammer...Treat everything as nail

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