Post Number:#2 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.