On this OEIS page, it is claimed that the number $2^n+3^n$ can never be a perfect square for positive integer $n$. Can you prove not only this claim, but also the following stronger statement?
$2^n+3^n$ can never be the square of a rational number for any (not necessarily positive) integer value of $n$.
The proof should be simple, involving only basic modular arithmetic and not more abstract concepts such as quadratic residues.
For $n=0$, we have $2^n + 3^n = 2$, which is not a square of any rational number.
We have $2^n + 3^n = p^2$.
Taking the equation mod 4, we have that the right-hand side is either 0 or 1, and the left-hand side is 1 or 3.
So to have equality $p^2 \equiv 2^n + 3^n \equiv 1 \pmod{4}$, which implies $n$ is even. Letting $s = \frac{n}{2}$ we get
$$(2^s)^2 + (3^s)^2 = p^2 \\ (3^s)^2 = p^2 - (2^s)^2 \\ (3^s)^2 = (p - 2^s)(p + 2^s)$$
In the right-hand side both factors must be powers of 3. If none of them is 1, both are multiples of 3 and then their difference, $2 \times 2^s$, is a multiple of 3 as well. But $2 \times 2^s$ is not a multiple of 3, so one of the factors is 1.
From $p - 2^s = 1$ and $p + 2^s = 3^{2s}$, subtracting both equations we get $$(3^s - 1)(3^s + 1) = 2^{s+1}$$ But the only two powers of 2 two units apart are 2 and 4, from where we must have $s = 1$. We can check that this value of $s$ does not satisfy our original equation, so there are no solutions.
We have $2^{-n} + 3^{-n} = \frac{p^2}{q^2}$, with $p$ and $q$ coprime. This implies
$$q^2\frac{2^n + 3^n}{2^n3^n} = p^2$$
The right-hand side is an integer, so the left one must be as well. Furthermore, $2^n + 3^n$ and $2^n3^n$ are coprime, so we must have $q^2 = k 2^n3^n$, where $gcd(k,p)=1$. Substituting back:
$$k(2^n+3^n) = p^2$$
$k$ and $p$ are coprime and $k$ is a factor of $p^2$, so $k=1$ and we have $2^n+3^n = p^2$, which we already showed is impossible.
x or y is zero
If x is zero, then $3^y + 1 = p^2 \Rightarrow 3^y = (p-1)(p+1)$. As the only powers of 3 two units apart are 1 and 3, $p=2$ and we have $\boxed{2^0 + 3^1 = 2^2}$.
If y is zero, then $2^x = (p-1)(p+1)$. As the only powers of 2 two units apart are 2 and 4, $p=3$, and we have $\boxed{2^3 + 3^0 = 3^2}$
x and y have the same sign
If x and y are positive, then by the same reasoning as earlier we can set $x = 2s$ and $y = 2t$ to get $3^{2t} = (p-2^s)(p+2^s) \Rightarrow (3^t-1)(3^t+1) = 2^{s+1}$ and because the only powers of 2 differing by 2 are 2 and 4, we have that the only solution is $s=2, t=1$. From this we get $\boxed{2^4 + 3^2 = 5^2}$.
From our earlier deduction, we know that the case in which both are negative is deeply intertwined with the case in which both are positive, and $-x,-y$ can only be a solution if $x,y$ is a solution. And in fact, our only positive solution also works for the negative case: $\boxed{2^{-4} + 3^{-2} = \left(\frac{5}{12}\right)^2}$
x is positive and y is negative
Assuming $x$ is positive and $y$ is negative, we have:
$$\frac{2^x3^y + 1}{3^y} = \frac{p^2}{q^2} \\ 2^x9^t + 1 = p^2 \\2^x9^t = (p-1)(p+1)$$ From this, either $p-1=2^s$ or $p+1=2^s$.
$p-1 = 2^s$:
$$2^s(2^s+2) = 2^x9^t \\ 2^{s-1} + 1 = 9^t \\ 2^{s-1} = (3^t-1)(3^t+1)$$ The only powers of two 2 units apart are 2 and 4, from where $s=4, t=1$ or $\boxed{2^5 + 3^{-2} = \left(\frac{17}{3}\right)^2}$.$p+1 = 2^s$:
$$2^s(2^s+2) = 2^x9^t \\ 2^{s-1} - 1 = 9^t$$ $2^{s-1}$ is 1 mod 3, from which $s-1$ is even. Say $s-1=2a$, then we have $$(2^a-1)(2^a+1) = 9^t$$ The only powers of three 2 units apart are 1 and 3, but then their product is not a power of 9. So this case is impossible.x is negative and y is positive
Doing the same thing for $x$ negative and $y$ positive, we arrive at $$4^t3^y + 1 = p^2 \\ 4^t3^y = (p-1)(p+1)$$ and analyzing the same two cases we find the final solution $\boxed{2^{-4} + 3^1 = \left(\frac{7}{4}\right)^2}$. Those six are the only solutions possible.
Consider the sequence:
$$ a_n=x^n+y^n $$
We can write this as a recurrence relation:
$$ \begin{align} a_n&=(x+y)\ a_{n-1}-xy\ a_{n-2} \\ a_n&=(x+y)\left(x^{n-1}+y^{n-1}\right)-xy\left(x^{n-2}+y^{n-2}\right) \\ a_n&=x^n+xy^{n-1}+x^{n-1}y+y^n-x^{n-1}y-xy^{n-1} \\ a_n&=x^n+y^n \end{align} $$
We are given $x=2,\ y=3$, so:
$$ \begin{align} a_n&=5a_{n-1}-6a_{n-2} \\ a_0&=2 \\ a_1&=5 \end{align} $$
Now consider the sequence modulo $15$:
$$ \begin{align} a_0&=2 \\ a_1&=5 \\ a_2&=13 \\ a_3&\equiv 5 \\ a_4&\equiv 7 \\ a_5&\equiv 5 \\ a_6&\equiv 13 \end{align} $$
Since the recurrence relation depends only on the two previous terms, and the two terms $a_5$ and $a_6$ are the same as $a_1$ and $a_2$, the sequence (modulo $15$) must repeat the four terms $5,13,5,7$ infinitely.
Now consider the sequence of squares,
$$ b_n=n^2 $$
Again we write this as a recurrence relation:
$$ \begin{align} b_n&=3b_{n-1}-3b_{n-2}+b_{n-3} \\ b_n&=3(n-1)^2-3(n-2)^2+(n-3)^2 \\ b_n&=3n^2-6n+3-3n^2+12n-12+n^2-6n+9 \\ b_n&=n^2 \end{align} $$
...and consider the sequence modulo $15$. This time it goes for quite a bit longer before repeating:
$$ b_n=0,1,4,9,1,10,6,4,4,6,10,1,9,4,1,0,1,4\ldots $$
But like before, once the initial three terms come up again it must repeat the same sequence infinitely.
Now notice that all elements of $a_n\bmod 15$ are in $\{2,5,7,13\}$; and all elements of $b_n\bmod 15$ are in $\{0,1,4,6,9,10\}$. Since these two sets are disjoint, no member of $a_n$ can be a member of $b_n$; or, no number of the form $a_n=2^n+3^n$ can be a square (for nonnegative $n$).
(Work in progress; negative powers to come later)
Given that the statement holds true for integers and that $2^n + 3^n$ always results in an integer the question can be answered by proving this stronger claim:
The square root of an integer is never a non-integer rational number
For the proof of that I like to quote this answer here https://math.stackexchange.com/q/4468:
Theorem: If $a$ and $b$ are positive integers, then $a^{1/b}$ is either irrational or an integer.
If $a^{1/b}=x/y$ where $y$ does not divide $x$, then $a=(a^{1/b})^b=x^b/y^b$ is not an integer (since $y^b$ does not divide $x^b$), giving a contradiction.
I subsequently found a variant of this proof on Wikipedia, under Proof by unique factorization.
The bracketed claim is proved below.
Lemma: If $y$ does not divide $x$, then $y^b$ does not divide $x^b$.
Unique prime factorisation implies that there exists a prime $p$ and positive integer $t$ such that $p^t$ divides $y$ while $p^t$ does not divide $x$. Therefore $p^{bt}$ divides $y^b$ while $p^{bt}$ does not divide $x^b$ (since otherwise $p^t$ would divide $x$). Hence $y^b$ does not divide $x^b$.
