## Equality and square

For discussing Olympiad Level Number Theory problems
Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Equality and square

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

Atonu Roy Chowdhury
Posts: 63
Joined: Fri Aug 05, 2016 7:57 pm

### Re: Equality and square

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.
This was freedom. Losing all hope was freedom.

Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

### Re: Equality and square

Thanks Antonu! 