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How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$?

This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is always bigger than the latter. This means I only have to account for the positive factors of $60$, and so I had the following pairs $(60,1), (30,2), (20,3), (15,4), (12,5)$, and I plugged them in and solved the systems. Of the 5 possible cases, only $(30,2)$ worked and yielded $x=4, y=14$.

Am I right in having done so? They don't have an answer, and when I tried to graph this, it didn't exactly work out well (I didn't get an integer value for $x$).

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    $\begingroup$ What about $(10, 6)$? :-) $\endgroup$ – Peter Košinár Sep 4 '17 at 17:57
  • $\begingroup$ Well $2^{2x} = 4^x \Rightarrow 4^x > 60 \Rightarrow x \geqslant 3$. We also have $-y^2 = 60 - 4^x = -(4^x - 60) \Rightarrow y^2 = 4^x - 60 \Rightarrow y^2 - 1 = 4^x - 61 \Rightarrow (y + 1)(y - 1) = 4^x - 61$ which means that $4^x - 61$ cannot be prime, so solutions exist if $x$ is even, and only some exist if $x$ is odd. Now it is a little easier to find solutions $\endgroup$ – Mr Pie Oct 22 '17 at 23:55
  • $\begingroup$ And from that, we see that $5\mid 4^{2n} - 61$ if $x = 2n$ and $3\mid 4^{2n - 1} - 61$ if $x = 2n - 1$, all for some $n \in \mathbb{N}$. And since $4^x$ is even and $61$ is odd, then $y^2$ is even hence $y$ is also even $\endgroup$ – Mr Pie Oct 23 '17 at 0:02
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You were so close. The last pair is $(10,6)$. That means $2^x=8$ or $x=3$ and $y=2$

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No you are not right.
You have neglected the case $\{ 2^x-y, 2^x+y \} = \{ 6, 10 \}$ ; which gives you $(x,y)=(3,2)$.



Notice that:

  • $(2^x+y)$ and $(2^x-y)$ have the same pairity;

  • product of $(2^x+y)$ and $(2^x-y)$ is even; so at least one of them is even;

  • by the above two remarks it follows that both of $(2^x+y)$ and $(2^x-y)$ are even.


  • So we can rewrite the equation as $\dfrac{2^x+y}{2} \cdot \dfrac{2^x-y}{2} = 15$ ;

  • the only possibilitis for $\{ \dfrac{2^x-y}{2}, \dfrac{2^x+y}{2} \}$ is $\{ 1, 15 \} $ and $\{ 3, 15 \} $ .

  • So we can conlude that:
    the only possibilitis for $\{ \ 2^x-y, \ 2^x+y \ \} $ is $\{ 2, 30 \} $ and $\{ 6, 10 \} $ .
    The first possibility gives $(x,y)=(4,14)$ and the second gives you $(x,y)=(3,2)$ .

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    $\begingroup$ It's given that $x, y$ are positive. $\endgroup$ – user263326 Sep 4 '17 at 20:36
  • $\begingroup$ @user263326 ; yes you are right. $\endgroup$ – Davood KHAJEHPOUR Sep 4 '17 at 20:51
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    $\begingroup$ Since $x, y$ are positive, the first possibility is $(x, y) = (4, 14)$ and the second is $(x, y) = (3, 2)$. $\endgroup$ – N. F. Taussig Sep 5 '17 at 9:24
  • $\begingroup$ @N. F. Taussig ; yes you are right. $\endgroup$ – Davood KHAJEHPOUR Sep 5 '17 at 9:26
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Basic things to note about $(2^x + y)(2^x - y) = ab = 60; a = 2^x + y; b=2^x-y)$.

1) $2^x + y > 0$ so $2^x - y > 0$.

2) $a= 2^x + y > 2^x -y=b > 0$ so $a > \sqrt{60} > 7$ and $b < \sqrt{60} < 8$.

3) $a$ and $b$ can't both be odd so they are both even.

So $b = 2,\color{red}4,6$ and $a=30,\color{red}{15},10$

4) $a+b = 2^{x+1} = 32, 16$ so $x=4,3$ and $y = \frac {a-b}2 = 14, 2$.

So the two possible solutions are $(x,y)=(4,14): (2^4+14)(2^4 -14)=30*2=60$ and $(x,y)=(3,2): (2^3+2)(2^3-2)=10*6=60$.

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$$(x,y) \in \lbrace(4,14),(3,2),(4,-14),(3,-2)\rbrace$$

Indeed, it's a difference of two squares:

$$2^{2x}-y^2=60$$ $$\text{let} \ \ \ \ t=2^x \ \ \implies $$ $$t^2-y^2= 60$$

In using the identity: $$\left(\frac{d_1+d_2}{2}\right)^2-\left(\frac{d_1-d_2}{2}\right)^2=d_1\cdot d_2$$

the only catch is that the pairs of divisors you choose need to be of the the same pairity, to allow that their sum or difference is divisible by two. Moving on, we have:

$$t=\frac{d_1+d_2}{2} \qquad \text{and} \qquad y= \frac{d_1-d_2}{2}$$

The total number of divisors of a given natural N will be:

$$\text{if} \ \ \ \ N=\prod_{i=1}^M p_i^{\alpha_i} \ \ \ \ \text{for finitely large M, and the i-th prime p}$$

$$\text{then} \ \ \ \ \left| \lbrace \text{set of divisors of N} \rbrace\right|=\prod_{i=1}^M (\alpha_i +1)$$

So for any number you factor it completely, and multiply the powers increased by one to get the amount of divisors of that number.

We now observe that $$60=2^2 \cdot 3 \cdot 5$$ $$(2+1)(1+1)(1+1)=12$$

So the divisors of 60 must include 12 numbers, starting from 1 it's not hard to find that they are:

$$\lbrace 1,2,3,4,5,6,10,12,15,20,30,60 \rbrace$$ And that the set of divisor-pairs, with each divisor of the same parity is only a two member set, so:

$$(d_1,d_2)\in \lbrace(30,2),(10,6)\rbrace$$

Thus

$$t=2^x=\frac{30+2}{2} \ \ \ \lor \ \ \ \frac{10+6}{2}$$ $$\text{so} \ \ \ \ t=16 \ \ \ \ \lor \ \ \ 8 \ \ \ \implies x=4 \ \ \ \lor \ \ \ 3$$ While $$y=\frac{30-2}{2}\ \ \ \lor \ \ \ \frac{10-6}{2} \implies y=14 \ \ \lor \ \ \ 2$$

Answers are then $$(x,y)=(4,14) \lor (3,2)$$ But we can see that we can let y be positive or negative, and not so for x, giving our above answer of 4 ordered pairs

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