Question

What are both pairs of integers (x,y) for 4^y-615=x^2.
Please show the steps!

Answer 1

\[
4^y -615 = x^2
\]This?

Answer 2

yes

Answer 3

Well, consider that \[
4^y−615=x^2 \\
\sqrt{4^y−615}=\sqrt{x^2} \\
\sqrt{4^y−615}=|x| \\
\pm \sqrt{4^y−615}=x \\
\]So if \((x, y)\) is a pair, then so is \((-x, y)\).

Answer 4

My next suggestion is to find a prime factorization of \(615\).

Answer 5

\(615= 3 \cdot 5\cdot 41\)

Answer 6

You want \(4^y - 3\cdot 5 \cdot 41\) to become a perfect square somehow.

Answer 7

o, now we can plug in numbers for y? Trial and error?

Answer 8

Hmmm, well I'm not sure of any mathmatical method for it other than trial and error right now but...

Answer 9

\[
4^y−615=x^2 \\
\log_4{(4^y - 615)} = \log_4(x^2) \\
\log_4{(4^y)} / \log_4{(615)}= \log_4(x^2) \\
y\cdot \log_4{(4)} = 2\log_4(x) \cdot \log_4{(615)} \\
y = 2\log_4(x) \cdot \log_4{(615)} \\
\]
Another way of looking at the problem

Answer 10

Hmmm, haven't they told you any methods for problems like these in your class?

Answer 11

Well, this chapter is on factoring difference of squares and cube, but i cant find the connection

Answer 12

Also i am a little confuse with the second method because i haven't learn log yet.

Answer 13

Are you sure it has \(4^y\)? That doesn't make sense then.

Answer 14

\(4y\) would make more sense.

Answer 15

it is 4^y

Answer 16

There is a method is am thinking of that worth trying. Which i learned in class to solve other problems in this chapter.
\[4^{y}=2^{2y}=2y^{2}\]
\[2y^{2}-x^{2}=615\]
\[(2^{y}+x)(2^{y}-x)=615\]

Answer 17

factors of 615: 1,615
3,123
5,41

Answer 18

\[2^{y}+x=41\] and \[2^{y}-x=5\], we will be able to find the value for y, right?

Answer 19

But y won't be an integer

Answer 20

Hmmm, let my try something... \[
\begin{array}{rcl}
4^y -615 &=& x^2\\
2^{2y} -615 &=& x^2\\
(2^y)^2 &=& x^2 +615\\
(2^y)^2 -x^2&=& 615\\
(2^y+x)(2^y -x)&=& 615\\
\end{array}
\]

Answer 21

@yociyoci Make sense?

Answer 22

I thought of this too but if we set the two binomials to be equal to the set of factors of 615 and then solve for y, y won't be an integer