A number theory problem by Morgan Blake

If $x = 0$ , $y^{2} = 8$ which has no solutions.

If $x = 1$ , $y^{2} = 9$ gives you $(0,3), (0,-3)$ .

Suppose $x \geq 2$ , LHS must be odd, so RHS must be odd too.

Let $y = 2k+1$ for some $k$ .

So that $2^{x} + 7 = (2k+1)^{2}$

$2^{x} = 4k^{2} + 4k - 6$

$2^{x-1} = 2k^{2} + 2k - 3 = 2(k^{2} + k) - 3$

LHS is even but RHS is odd, so there are no solutions left.

Number of solutions = $\boxed{2}$ . ~~~

We take the equation modulo 4, since squares are always $0$ or $1$ modulo 4, and because $2^x$ works conveniently with modulo 4.

If $x>1$ , $2^x+7=y^2\implies 3\equiv y^2\pmod 4$ which has no solutions. If $x=1$ , $2^1+7=y^2\implies y=-3,3$ If $x=0$ , $8=y^2$ , which has no solutions. $x$ cannot be negative, because then the LHS is not an integer.

The answer is $\boxed{2}$ .

For $x\geq 2$ , we can apply residue test modulo $4$ . Since $4$ has only $2$ distinct quadratic residues: $0$ and $1$ , there's no solution for $x\geq 2$ because, $2^x+7\equiv 3 \not\in \{0,1\}\pmod{4}$ . Just check $0$ and $1$ and we can find that $x=1$ works and it gives $y=\pm 3$ , that's $\fbox{2}$ ordered pairs.

At first the equation $2^{x}+7=y^{2}$ looks quite intimidating, but with a bit of thought, we realise that perhaps modular arithmetic is the best way to go.

The two obvious moduli don't really allow us to gain any ground (mod 7 and mod y), so we'll have to be a bit smarter.

Working with modulus 2 looks promising but doesn't quite work. Instead, we can try mod 4.

This yields: $0+3 \equiv y^{2} \pmod{4} \text{ when } x>1$ $2+3 \equiv y^{2} \pmod{4} \text{ when } x=1$ $1+3 \equiv y^{2} \pmod{4} \text{ when } x=0$

$y^{2} \equiv 0 \pmod{4} \text{ or } y^{2} \equiv 0 \pmod{4}$ , ( $y^{2} \not= 3 \pmod{4}$ )

Therefore, x can only be 0 or 1.

However, if $x=0$ then $y^{2}=8$ , to which the solution for y is obviously not an integer.

Hence, $x=1$ , so $y=3$ or $y=-3$

So, the only ordered solutions are $(1,-3) \text{ and } (1,3)$ .

There are two ordered solutions, so the answer is $\boxed{2}$

Not really sure if this is a valid solution.

We can rearrange the equation to:

$y^2 - 2^x = 7$

Using difference of two squares, this can be factorised.

$(y - 2^\frac{x}{2} )\times (y + 2^\frac{x}{2} ) = 7$

Now, let us consider factors.

$7 = 1 \times 7$

&

$-1 \times -7$ .

Since the two factors for 7 above are different, and it is clear that one bracket cannot equal both -7 and -1, we can see that the number of solutions is $\boxed {2}$

If x>1, then the left side $y^2\equiv 3\pmod{4}$ ,which is impossible, because squares are congruent to 0 or 1 modulo 4.So x=2 and $y=\pm3$