### Equality and square

Posted:

**Fri Aug 04, 2017 1:24 pm**Determine all pairs $(a, b)$ of integers such that

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

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

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Posted: **Fri Aug 04, 2017 1:24 pm**

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

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

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

Posted: **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.

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.

Posted: **Sat Aug 19, 2017 11:09 pm**

Thanks Antonu!