Equality and square

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Katy729
Posts: 40
Joined: Sat May 06, 2017 2:30 am

Equality and square

Unread post by Katy729 » Fri Aug 04, 2017 1:24 pm

Determine all pairs $(a, b)$ of integers such that
$1+2^{a}+2^{2b+1}= b^{2}$

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Atonu Roy Chowdhury
Posts: 40
Joined: Fri Aug 05, 2016 7:57 pm
Location: Chittagong, Bangladesh

Re: Equality and square

Unread post by Atonu Roy Chowdhury » Sat Aug 05, 2017 11:54 pm

The case of negative integers is quite trivial.
Now we'll work with case $a,b > 0$

Lemma 1: $x < 2^x$
Proof: We'll prove it by induction. Base case is solved.
Now assume $x<2^x \Rightarrow x+1<2^x+1 \le 2^{x+1}$. Done!

Lemma 2: $b^2 < 2^{2b+1}$
Proof: We'll prove it by induction. Base case is solved.
Now assume $b^2 < 2^{2b+1}$. Lemma 1 gives us $2b+1 < 2^{2b+1} < 2^{2b+1} (4-1) = 2^{2b+3}-2^{2b+1}$
Summing up the two ineqs, we get $(b+1)^2 < 2^{2b+3}=2^{2(b+1)+1}$

No need to go back to our problem. Our problem is solved. No such pair exists.

Katy729
Posts: 40
Joined: Sat May 06, 2017 2:30 am

Re: Equality and square

Unread post by Katy729 » Sat Aug 19, 2017 11:09 pm

Thanks Antonu! :)

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