# Is it possible that $2^{2A}+2^{2B}$ is a square number?

Let A and B be two positive integers greater than $$0$$. Is it possible that $$2^{2A}+2^{2B}$$ is a square number?

I am having trouble with this exercise because I get the feeling the answer is no, but I cannot elaborate on the proof. So far what I thought was to assume that there is some integer $$C>0$$ such that $$2^{2A}+2^{2B}=C^2$$. Then $$(2^A+2^B)^2=C^2+2^{A+B+1}$$ I was trying to see if the previous expression could hold a contradiction but I got stuck. All I could find is that $$C$$ needs to be an even number but that doesn't seem to get me anywhere. I'd appreciate any help.

• Can you prove that $4^{A-B}+1$ is never a square? Aug 15, 2020 at 16:38
• Is there a certain reason you wrote "even powers of 2" instead of "powers of 4"? (At first I thought you meant "powers of 2 which are even numbers," or, in other words, "powers of 2 other than 1.") Aug 16, 2020 at 4:25
• You were close to something when considering $C$ being even, that is basically looking at the equation modulo $2$. Turns out, a really simple solution arise when you look at it modulo $3$ (see below).
– Sil
Sep 5, 2020 at 7:04

Without loss of generality, let $$A>B$$. Then $$2^{2A}+2^{2B}=2^{2B}(2^{2A-2B}+1)$$ is a square implies $$2^{2A-2B}+1$$ is a square as $$2^{2B}$$ is a square. But this is impossible since $$2^{2A-2B}$$ is a square.

Shubhrajit Bhattacharya's answer gives a simple, direct proof that $$2^{2A}+2^{2B}$$ cannot be a square. But just for fun, let's finish off the OP's approach (which I initially thought led to a dead end).

If $$(2^A+2^B)^2=C^2+2^{A+B+1}$$, then $$(2^A+2^B+C)(2^A+2^B-C)=2^{A+B+1}$$, which means that $$2^A+2^B+C$$ and $$2^A+2^B-C$$ are both powers of $$2$$, and obviously different powers of $$2$$, say $$2^a$$ and $$2^b$$ with $$a\gt b$$ and $$a+b=A+B+1$$. But this implies

$$2(2^A+2^B)=2^a+2^b$$

If we now assume, without loss of generality, that $$A\ge B$$, we have

$$2^{B+1}(2^{A-B}+1)=2^b(2^{a-b}+1)$$

Now $$a\gt b$$ implies $$2^{a-b}+1$$ is an odd number greater than $$1$$, from which it follows that we must have $$A\gt B$$ (otherwise the left hand side is a power of $$2$$, not a multiple of an odd number greater than $$1$$). This in turn implies $$b=B+1$$ and $$a-b=A-B$$, from which we get

$$a+b=(a-b)+2b=(A-B)+2(B+1)=A+B+2$$

in contradiction to $$a+b=A+B+1$$.

Remark: I was a little surprised by the nature of the contradiction here, and had to check my work carefully to make sure I hadn't made a stupid arithmetic mistake.

Just do it.

Assume without loss of generality that $$A \le B$$ so

$$2^{2A} + 2^{2B}=$$

$$2^{2A} (1 + 2^{2B-2A})=$$

$$(2^A)^2 [1 + 2^{2B-2A}]=$$

$$(2^A)^2 [(2^{B-A})^2 + 1]$$.

So if that is a perfect square then we must have $$(2^{B-A})^2 + 1$$ being a perfect square.

But $$(2^{B-A})^2$$ is a perfect square so we have two consecutive perfect squares. It should be easy to convince yourself that the only time that ever occurs is $$0^2$$ and $$1^2$$. (Proof as addendum).

So the only way this can happen is if $$(2^{B-A})^2 = 0$$ and $$(2^{B-A})^2 + 1=1$$.

But $$2^{B-A} = 0$$ is not possible.

====

Addendume: Then only two consecutive squares are $$0$$ and $$1$$.

Proof: Suppose $$m^2 = n^2 + 1$$. where $$m,n$$ are non-negative integers. $$n^2 < m^2 = n^2 + 1 \le n^2 + 2n + 1= (n+1)^2$$ so $$n < m \le m+1$$. But the only integers between $$n$$ (exclusive) and $$n+1$$ (inclusive) is $$n+1$$ so $$m = n+1$$. And so $$n^2 + 1 = m^2 = (n+1) = n^2 + 2n + 1$$ so $$2n = 0$$ and $$n = 0$$ and $$m =1$$.

Assume that $$2^{2A}+2^{2B}$$ is a perfect square. Without loss of generality, assume $$A \geqslant B$$. Then, let $$A-B=x$$, where $$x$$ is a non-negative integer. It follows that we have: $$2^{2A}+2^{2B}=(2^B)^2 \cdot (2^{2x}+1)$$ Now, if the LHS is a perfect square, then the RHS must also be a perfect square. It follows that $$2^{2x}+1$$ is a perfect square. Let this be $$n^2$$. We then have: $$2^{2x}=n^2-1=(n-1)(n+1)$$ Now, we need $$n-1$$ and $$n+1$$ to both be perfect powers of $$2$$. This can only happen for $$n=3$$. However, even then, we would only have $$2^{2x}=8$$ which is impossible as $$x$$ is an integer. Thus, no solutions exist.

• "Now, we need n−1 and n+1 to both be perfect squares. This can only happen for n=3." — Is 3-1 a perfect square? Aug 15, 2020 at 16:45
• @MarianD sorry, that was supposed to be perfect powers of 2. Not perfect squares. I have made the edit. Aug 16, 2020 at 16:20
• Oh no! Your original sentence “It follows that $2^{2x}+1$ is a perfect powers of $2$.” was correct (not grammatically, “power” instead of “powers” :-)). Your corrected sentence is not true, sorry. Aug 16, 2020 at 17:30
• @MarianD Lol, sorry about that, I was a bit hasty and made the edit in the wrong place. Should be fine now. Aug 16, 2020 at 18:18
• Yes, it is OK now. Aug 16, 2020 at 18:22

We would have $$k^2=4^{A}+4^{B}\equiv 1+1= 2\pmod 3$$, impossible as $$k^2 \equiv 0,1 \pmod 3$$.