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/sci/ - Science & Math

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1763263 No.1763263 [Reply] [Original]

Find all ordered pairs of integers (x; y) such that x2 + 2x + 18 = y2.

>> No.1763268

no

>> No.1763266

NO U

>> No.1763272

plz

>> No.1763276

(x+1)^2 + 17 = y^2
(x+1)^2 = y^2 - 17

So on the right hand side you're looking for a perfect square which is 17 more than some other perfect square. Now the difference between successive perfect squares is successive odd integers. 17 is the 9th odd number so we're looking for the 9th perfect square, 81. (8^2 = 64 = 9^2 - 17 = 81 - 17). Thus y = +- 9, which gives (x+1)^2 = 64, so x = 7 or -9.

y = +-9
x = 7 or -9

>> No.1763277

(x+1)^2+17=y^2
(x+1)^2-y^2=17
(x-y+1)(x+y+1) = 17

case 1
x-y+1 = -1
x+y+1 = -17
x = -8
y = -6

case 2
x-y+1 = -17
x+y+1 = -1

(x,y) = (-8,6)

case 3
x-y+1 = 1
x+y+1 = 17
etc.

>> No.1763279

(-9,+-9), (7,+-9)

sauce on that pic

>> No.1763280

>>1763277
i effed that up, but that's the idea.

>>1763276
works, but inefficient and clumsy.

>> No.1763282

>>1763280
>Ineffecient
>He has to solve 4 systems of linear equations, while I make an argument from number theory.
>laughing girls.jpg

>> No.1763290

>>1763282

If you're going to be cocky about it, I'll point out that you never justified why (x+1) and y must be consecutive squares.

Your "solution" fails when the equation is

x^2 + 2x + 16 = y^2

>> No.1763340

>>1763290
Uhm because the left hand side (x+1)^2 is an integer and a perfect square, then since the right hand side y^2 - 17 is EQUAL to the left hand side it is ALSO an integer and a perfect square. y^2 is an integer and a perfect square. Ok, why they have to be consecutive, maybe I didn't justify that enough. But if you look at the differences between perfect squares they are of the form 2n+1. 1 3 5 7 9 11 13 15 17. Find any sub-sequence that that sums to 17.

Lrn2 natural numbers?

>> No.1763351

>>1763290
(x+1)^2 = y^2-16

Perfect square which is 16 more than some other perfect square... Well just listing the first perfect squares 1 4 9 16 25 36 49 I can see that it has to be 25 and 9.

>> No.1763366
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1763366

>>1763351

>> No.1763627

>>1763351
>>1763366
Or I can just
a^2 + 4^2 = b^2
Oh where have I seen something like this before... Maybe a 345 triangle when I learned about the Pythagorean theorem!

>> No.1763676

>>1763627
>when I learned about the Pythagorean theorem

>> No.1763770

>>1763340

>But if you look at the differences between perfect squares they are of the form 2n+1

Really? 4^2 - 2^2 = 12.

12 = 2n + 1???

>Lrn2 natural numbers?

Take your own advice.

>> No.1763808

>>1763351

>(x+1)^2 = y^2-16

Reread the equation

x^2 + 2x + 16 = y^2

(x+1)^2 = y^2 - 15.

My point was that by the original "solution" we'd have to look for consecutive squares which would give us y^2 = 8^2 and (x + 1)^2 = 7^2.

The problem is that you'd be missing y^2 = 4^2 and (x + 1)^2 = 1^2.

So the reply seems to be, just find perfect squares whose difference is what you're looking for. But what happens when the difference isn't a small number. Say it's 120? How tedious is it going to be to search through perfect squares.

No, this anon had it right: >>1763277

For the problem above

(y + x + 1)(y - x - 1) = y^2 - (x + 1)^2 = 15 = (1)(15) = (3)(5) = (-1)(-15)=(-3)(-5)

So (y + x + 1) + (y - x - 1) = 1 + 15 or 3 + 5 or -1 + -15 or -3 + -5

In other words y is in the set {-8, -4, 4, 8} and from there it's rather easy to find the corresponding x values.

This method generalizes completely if

x^2 + 2x + 121 = y^2 or (y + x + 1)(y - x - 1) = 120

You just need to look at all of the divisors of 120 in I and then simply go through the same algorithm.

>> No.1763825

>>1763808
>Say it's 120?
2n+1 + 2n +3 = 4n + 4 =120, n = 116/4 = 29
31^2 - 29^2 = (31+29)(31-29) = 60*2 =120

>> No.1763833

>>1763770
>Really? 4^2 - 2^2 = 12.
Obviously between consecutive ones, asshat.

>>1763808
You're right, I derped so hard I herped.

>> No.1763848

>>1763825
Hmm when I think more about it maybe this method doesn't generate all solutions unless you consider more possible squares.

You would need an expression for possible differences between squares and solve that. Not very hard, really. A bit messy and lengthy yes, but so's the simultaneous equation method.

>> No.1763864

>>1763825

What about 17^2 - 13^2? or 11^2 - 1^2 or 13^2 - 7^2?

>> No.1763876

>>1763848

Keep thinking about it and you'll realize that the solution you criticized is actually the best way to go about it. It's only as tedious as factoring and there's no way you're going to get around that.

>> No.1763901

>>1763864
Two steps: possible as shown before
Three steps: can't be done, sum of three odd numbers is odd.
Four steps
8n + 16 = 120
n = 104/8 = 13
17^2 - 13^2 = (17-13)(17+13) = 30*4 = 120
Six steps:
12n + 36 = 120
n = 84/12 = 7
13^2 - 7^2 = (13+7)(13-7) = 20*6 = 120
Eight steps:
16n + 64 = 120
n = 54/16, not an integer. Ignore.
Ten steps:
20n + 100 = 120
n = 1.
11^2 - 1^2 = 121-1 = 120.

So you're gonna have to iterate the method for each conceivable number of steps between the squares. Still simpler than simultaneous equations I think.

>> No.1763910

>all of the divisors of 120
>http://www.wolframalpha.com/input/?i=120+divisors
>all 32 of them
>fuck yeah.jpg

>> No.1763925

What class is this for?

>> No.1763964

>>1763901
Cycling through these
O(sqrt(n)) right? Complexity for finding all the divisors of an integer is O(sqrt(n)) too, right? But then you have to solve simultaneous equations too...

>> No.1763990

>>1763901
For 17 you have
One step: 2n +1 = 17, n = 8
9^2 - 8^2 = 17. Ok.
Three steps: 6n + 9 = 17
n = 8/6. Not integer.
Five steps: 6n + 25 = 17. Not solvable with naturals. Stop.

For 15
One step: 2n + 1 = 15
n = 14/2 = 7
8^2 - 7^2 = 15. Ok.
Three steps: 6n + 9 = 6
n = 1
4^2-1^2 = 15. Ok.
Stop.

>> No.1764007

3x^2+18=y^2
sqrt(3x)-sqrt(y)=-18
x=-2, y=-9

Easy maths brah, enjoy your A+ grade in University.

>> No.1764010

>>1764007
4/10.

>> No.1764019
File: 141 KB, 279x336, young caucasian male.png [View same] [iqdb] [saucenao] [google]
1764019

>>1764007

>> No.1764023

>>1764019

>young caucasian male.jpg

>> No.1764411

>>1763925

Linear algebra 135

>> No.1764448

>>1764023
.png

>> No.1765253

lol isnt this from university of waterloo, maple TA math 135 assignment 1? you cant solve such a simple question and going to uw...

>> No.1765269

woops its actually the written one not mapleta, same thing

>> No.1766531
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1766531

bump with rainbow-shitting plane

>> No.1766625

>>1765253
fucking LOL, I am in this class too. Such a small world ... people posting my homework assignments on /sci/.

>> No.1766652

>>1766625
Dude, math at Waterloo would have been AWESOME.

>> No.1766713

>>1763901

I know you're not around anymore since I've had a refreshing night's sleep, but for the short-term posterity that is 4chan...

What you're effectively doing is saying... find me k consecutive odd numbers starting at, say (2n + 1), such that they sum to 120.

That is, you want to solve <span class="math">\sum_{i = 1}^k 2n + (2i - 1) = 120[/spoiler].

Simplifying the sum, we have: <span class="math">2kn + k^2 = 120[/spoiler].

So in the end you're solving a quadratic Diophantine in both cases. You've got a method for doing that that simply runs through k. I suppose that this is fairly easy to code, but is somewhat inefficient.

The other method of simply factoring 120 is, I think, a much more established problem. By hand I think you're going to make fewer mistakes and by computer there are already tons of factorization libraries available.

Moreover, you don't need to deal with every two-term factorization of 120. Only cases where both terms have the same parity are relevant.

Finally, you keep talking about solving the equations as being difficult. But it's really not because of the form they take. Since (y + x + 1) + (y - x - 1) = 2y, once you have a factorization you just have to add the terms and divide by two to get y (this is why only like-parity factorizations matter). So 120 = (4)(30) and (4 + 30) / 2 = 17 is a possible y value. Getting from the factorization to the solutions is trivial.

>> No.1768207

lol well its not just your class, the assignments are the same for all sections.. theres like 10 classes, so yeah

>> No.1768210

btw im in wentang kuos 2:30pm class woot LOL

>> No.1768217

woops I meant 1:30 haha