$2^n a+b=x_n^2$ then $x_{n+2}^2-b=4(x_n^2-b)$ $x_{n+2}^2-(2x_n)^2=-3b$ but equation$x^2-y^2=c$ has finite solution.thus $x_k=x_l$ for some $k\not= l$ hence $a=0$

A proof from first principles. Let $(a,b)$ be a counterexample with smallest $|a|+|b|$. Theorem: b is odd. Proof 1. $b\ne 0$. For otherwise $a$ is a non-zero square with $2a$ square. This contradicts the irrationality of the square root of 2. 2. $b$ is not divisible by 4 If it were, then for all n, $2^na+\frac{b}4=\frac{2^{n+2}a+b}4$ would be square, so $(a,\frac{b}4)$ would be a smaller counterexample. 3. $b$ is odd Otherwise $4a+b$ is a square which is divisible by 2 but not 4. QED Then suppose that for all $n$, $2^na+b=t_n^2$ where $t_n\ge 0$. Theorem: For all $n$ with $2^{n-2}>|a|$ and $n>0$ that there exists some odd factor $x$ of $a$ such that $t_n=2^{n-2}y-x$ where $y=\frac{a}{x}$. Proof Note that $(t_{n+1}+t_n)(t_{n+1}-t_n)=t_{n+1}^2-t_n^2=2^na$. Furhter, since $b$ is odd, $t_n$ and $t_{n+1}$ are both odd. Thus both of $t_{n+1}+t_n$ and $t_{n+1}-t_n$ are even, but only one is a multiple of 4. Thus there exist $x,y$ with xy=a and with x odd such that $\{t_{n+1}+t_n,t_{n+1}-t_n\}=\{2x,2^{n-1}y\}$. However, $2^{n-1}y\ge 2^{n-1}>2|a|\ge 2x$. Thus $t_{n+1}+t_n=2^{n-1}y$ and $t_{n+1}-t_n=2x$. Subtracting gives the required expression for $t_n$. QED Theorem: For any odd factor x of a, there can be at most 2 n such that $(2^{n-2}y-x)^2=2^na+b$ where $y=\frac{a}{x}$. Proof To see that there are at most 2 such n, expand and multiply by 16 to get $2^{2n}y^2-2^{n+3}a+x^2=2^{n+4}a+b$. Letting $z=2^n$, this gives $y^2z^2-(24a)z+x^2-b=0$, which is a nondegenerate quadratic equation with at most 2 solutions. QED We have shown that for all sufficiently large n there is such an x. However, there are only finitely many such x, and each corresponds to at most 2 n. This contradiction shows that there are no counterexamples.

Can you write your solution a bit better, Hawk Tiger? If it is correct, it's very short.

See at http://www.mathlinks.ro/Forum/viewtopic.php?search_id=600084763&t=145117 http://www.mathlinks.ro/Forum/viewtopic.php?p=150403#150403

Probably this is what Hawk Tiger meant, written out. edriv wrote: For each n: $ 2^{n}a + b$ is a perfect square $ \Rightarrow 4(2^{n}a + b) = 2^{n + 2}a + 4b$ is a perfect square. Also $ 2^{n + 2}a + b$ is a perfect square. Then we have two perfect squares at distance $ 3b$. We con construct arbitrary large perfect squares with distance $ 3b$. But this is impossible, then: - the distance is 0, $ b = 0$, but then $ a,2a$ are both perfect squares $ \Rightarrow a = 0$ - then our squares are not arbitrary large: $ a = 0$.

Trying to solve this problem, i found some curious facts about sequences that satisfy the recursive equation $X_{n+2}=3X_{n+1}-2X_n$. It's easy to see that the sequence of $2^n a+b$ where $n$ spans the nonnegative integers satisfy the equation. It's also easy to prove that $X_n$ can be written as $\alpha_nX_1+\beta_nX_0$ where $\{\alpha_n\}$ and $\{\beta_n\}$ are also such sequences. Namely, $\alpha_n=2^n-1$ and $\beta_n=2-2^n$. Another curious fact is that $X_n \equiv X_{n+j} \pmod{\alpha_j}, j\geq 2$ Nevertheless, this facts didn't help me to solve the problem... ;(

There is an extremely nice generalization of this result: Let $W(x)$ be a polynomial with integer coefficients such that $W(2^{n})$ is a perfect square for all positive integers $n$. Then there exist another polynomial $Q(x)$ with integer coefficients such that $W(x)=Q(x)^{2}$

if $a$ is nonzero then $2^n{a}+b$ can be any big which implies : $x^2=2^n{a}+b , y^2=2^{n+2}{a}+b \implies 4x^2-y^2=3b$ which means we got infinite pair of positive integers like $(2x,y)$ such as that the difference of their squares is a constant which a contradiction...

Since $b$ is a quadratic residue $\pmod{2^n}$ for every $n$, it must itself be a perfect square. Let $b=c^2$, where $c>0$. Suppose $2^n a+b=d_n^2$ for $n\geq 0$. Note that this implies that $a\geq 0$. Suppose that $a\neq 0$. Then $$d_n+c=2^i k_1, \quad d_n-c=2^j k_2$$where $2\nmid k_1 k_2$, $a\geq k_1,k_2\geq 1$ and $i+j\geq n$. This implies that $$c=2^{i-1}k_1-2^{j-1}k_2 \qquad (1)$$Let $v_2(c)=t$ then $(1)$ implies that $\min(i,j)=t+1$. So for any $n$ either $$c=2^t k_1-2^{j-1}k_2< 2^t a-2^{n-t-2} \qquad \text{or} \qquad c=2^{i-1}k_1-2^t k_2 >2^{n-t-2}-2^t a.$$In any case if we take $n\to \infty$ we get a contradiction, as desired.