If x^2 - 5y^2 = 1232, how many pairs are possible for (x,y) ?

[Quant Warrior]

If x^2 - 5y^2 = 1232, how many pairs are possible for (x,y) ?

1) 2
2) 3
3) 1
4) 0

If there were no "5" before y^2, I could have tried to solve it by breaking x^2-y^2 into (x-y)(x+y).
Like this : (x-y)(x+y) = 1232 . At least I would have made an attempt. Buy I am at a loss as to how to go about that question.

Please enlightenment me with clarity on this.

 

[pka]

If x^2 - 5y^2 = 1232, how many pairs are possible for (x,y) ?
1) 2 2) 3 3) 1 4) 0
If there were no "5" before y^2, I could have tried to solve it by breaking x^2-y^2 into (x-y)(x+y).
Like this : (x-y)(x+y) = 1232 . At least I would have made an attempt. Buy I am at a loss as to how to go about that question.
Please enlightenment me with clarity on this.

What is your question? Have you studied the hyperbola?
See here.

 

[JeffM]

There are an infinite number of pairs of real numbers that satisfy this equation

[MATH]x^2 - 5y^2 = 1232 \implies x^2 = 1232 + 5y^2 \implies x = \pm \sqrt{1232 + y^2}. [/MATH]
So x can be any real number with an absolute value slightly above 35.

Or must x and y be integers?

EDIT: I do not see an obvious reason why there will be a finite number of solutions even if we are just considering integers. There is a sizable number of such pairs to review. My question about integers was simply looking to determine if there are one or more undisclosed restrictions on the solution.

 

[Dr.Peterson]

I'd guess that you did mean to restrict the variables to integers (or positive integers) and forgot to state that. (Shame on you!)

Are you studying Pell equations? What techniques have you learned?

 

[HallsofIvy]

If x^2 - 5y^2 = 1232, how many pairs are possible for (x,y) ?

1) 2
2) 3
3) 1
4) 0

If there were no "5" before y^2, I could have tried to solve it by breaking x^2-y^2 into (x-y)(x+y).
Like this : (x-y)(x+y) = 1232 . At least I would have made an attempt. Buy I am at a loss as to how to go about that question.

Please enlightenment me with clarity on this.

As has been pointed out, there are an infinite number of pairs of real numbers that satisfy that. If you mean that x and y must be positive integers then after you have (x- y)(x+ y)= 1232, look at the ways to factor 1232 into two integers.

1232= 2(616)= 4(308)= 8(154)= 16(77)= 16(7)(11)
Thats a total of 4+ 1+ 1= 6 prime factors.

Now it becomes "how many ways can 6 be written as a+ b for a and b positive integers?" 5+ 1= 4+ 2= 3+ 3= 6.
5+ 1= 6 gives [2^4(7)](11)= (112)(11)= 1232
4+ 2= 6 gives [2^3(7)][(2)(11)]= (56)(22)= 1232
or [2^3(11)][2(7)]= (88)(14)= 1232
3+3= 6 gives [2^2(7)][2^2(11)]= (28)(44)= 1232

 

[JeffM]

As has been pointed out, there are an infinite number of pairs of real numbers that satisfy that. If you mean that x and y must be positive integers then after you have (x- y)(x+ y)= 1232, look at the ways to factor 1232 into two integers.

1232= 2(616)= 4(308)= 8(154)= 16(77)= 16(7)(11)
Thats a total of 4+ 1+ 1= 6 prime factors.

Now it becomes "how many ways can 6 be written as a+ b for a and b positive integers?" 5+ 1= 4+ 2= 3+ 3= 6.
5+ 1= 6 gives [2^4(7)](11)= (112)(11)= 1232
4+ 2= 6 gives [2^3(7)][(2)(11)]= (56)(22)= 1232
or [2^3(11)][2(7)]= (88)(14)= 1232
3+3= 6 gives [2^2(7)][2^2(11)]= (28)(44)= 1232

I am confused.

[MATH]1232 = x^2 - 5y^2 = (x - y\sqrt{5})(x + y\sqrt{5}) [/MATH]
takes us right out of the realm of integers.

Sorry to be stupid.

 

[HallsofIvy]

Yes, I confused the simple example with the actual problem. I would be inclined to answer "4) 0".

 

[JeffM]

Yes, I confused the simple example with the actual problem. I would be inclined to answer "4) 0".

My ignorance of Diophantine equations is bottomless, but I agree. Here is my reasoning.

[MATH]5y^2 = 1232 + x^2 \implies y^2 = \dfrac{1232}{5} + \dfrac{x^2}{5} = 246 + \dfrac{x^2 + 2}{5}.[/MATH]
Now a number evenly divisible by 5 has a low order digit of 5 or of 0. That means that if y^2 is an integer, then the low order digit of x^2 is 3 or 8.
Assume for purposes of contradiction that there is such a perfect square.

[MATH]j = \sqrt{x^2} = 10k + d,[/MATH]
where k is a non-negative integer and d is a non-negative integer < 10.

[MATH]\therefore x^2 = 100k^2 + 20kd + d^2.[/MATH]
[MATH]\begin {array} {|c|c|} \hline d & d^2 \\\hline 0 & 0 \\\hline 1 & 1 \\\hline 2 & 2 \\\hline 3 & 9 \\\hline 4 & 16 \\\hline 5 & 25 \\\hline 6 & 36 \\\hline 7 & 49 \\\hline 8 & 64 \\\hline 9 & 81 \\\hline \end{array}[/MATH]
But there is no d that generates a square with a low order digit 2 or 8. Therefore there is no such x^2 and therefore no integer y^2. But that entails there is no integer y.

EDIT: This actually explodes my intuition. Actually I guess it should not. Either zero or infinite is what my intuition probably should have told me.