Determine all pairs $(a, b)$ of integers such that

$1+2^{a}+2^{2b+1}= b^{2}$

## Equality and square

- Atonu Roy Chowdhury
**Posts:**63**Joined:**Fri Aug 05, 2016 7:57 pm**Location:**Chittagong, Bangladesh

### Re: Equality and square

The case of negative integers is quite trivial.

Now we'll work with case $a,b > 0$

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!

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.

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.

This was freedom. Losing all hope was freedom.