Solving a Quadratic Diophantine Equation

This video is about a quadratic diophantine equation.
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#ChallengingMathProblems #DiophantineEquations

2 BOOKS and SOME LINKS:

[SyberMath]

Diophantine Equations are fun to solve because the number of variables is greater than the number of equations. There are many kinds of Diophantine Equations and a decent number of books written on them. This can be considered an easyish problem since the solution depends on a simple idea. Divisibility and modular arithmetic also play an important role in solving these equations. The linear ones are the easiest ones but even those have interesting results! Any thoughts? Plz share here

[luggepytt]

Here is how I did it:

First, move the 9 to the right-hand side and factor both sides:
x²+3x = 9y²-9
x(x+3) = 9(y-1)(y+1)

Since the right-hand is divisible by 3, then 3 must divide x or (x+3). And in fact, if it divides one of them, then it divides both.
So let’s substitute 3z = x:
3z(3z+3) = 9(y-1)(y+1)

Divide both sides by 9:
z(z+1) = (y-1)(y+1)

We can easily make the left and right side equal by making them both equal to zero, i.e.
z = 0 or z = -1
and
y = -1 or y = 1

After substituting back x = 3z we get the four solutions:
(x, y) in {(0, -1), (0, 1), (-3, -1), (-3, 1)}


Are there other solutions? We have to investigate three cases:

1. y = 0: This does not work. z(z+1) = -1 has no integer solution.

2. y > 1: Look at the right-hand side, (y-1)(y+1). It is a product of two positive numbers whose difference is 2. Then look at the left hand side, z(z+1).
2.a. If z is positive, the left-hand side is the product of two positive numbers whose difference is 1. This means that z and z+1 must both lie between y-1 and y+1, in other words,
y-1 < z < z+1 < y+1. But it is impossible to squeeze in two different integers between y-1 and y+1, there is only room for one (namely y). So z cannot be an integer.
2.b. If z is negative, set w = -z, and the left-hand side becomes (-w)(-w+1) = w(w-1). Then apply the same argument as in 2.a.

3. y < -1: Analogous argument as in 2.

[hkemal]

Did you hear that old math teachers never die? They just lose some of their functions.

[pierreneau]

Nice demonstration. Other way transform the equation in
(2x+3-6y)(2x+3+6y)=-27

[ManuelRuiz-xibt]

Completing the square for x, moving 9y² to the left hand side, and remaining constants to the right, yields
x² + 3x + 9/4 - 9y² = -9 + 9/4
(x + 3/2)² - 9y² = -9 + 9/4
(2x + 3)² - 36y² = -36 + 9
(2x + 3 + 6y)(2x + 3 - 6y) = -27
Only integer factors of -27 are +/-1, +/-3, +/-9 and +/-27.

So the differnce of the factors 12y is in {-26, -12, 12, 26}.
Hence y = +/-1 and x= 0.

[christianthomas]

I found the same solutions like this:
x(x+3) = 9(y**2-1)
either x=0 or x=-3
for x=0, y=+/-1
for x=-3, y=+/-1, then the four integer solutions

[SuperYoonHo]

Why would anyone say they don't like this fabulous video?!! If they really do so, they have not realized a math genius underground🤔

[mohammadshaanmdsalamsaikh]

REALLY WONDERFUL, SIR. I JUST APPRECIATE THE WAY YOU SOLVE ANY PROBLEM... PLEASE KEEP POSTING VIDEOS ON OLYMPIAD PROBLEMS...
AND I HAD A REQUEST TOO... I DIDN'T FIND ANY VIDEO POSTING SOLUTIONS ON THE BOOK "PATHFINDER FOR OLMPIAD MATHEMATICS". IF POSSIBLE
.. PLEASE POST VIDEO SOLUTIONS OF THAT BOOK... THANKS A LOT

[mcwulf]

x must be a multiple of 3 so put x=3u. Eqn becomes, after dividing by 9:

u^2 + u + 1 = y^2
(u+1) = (y+u)(y-u)

Only divides if y = +/-1.

[mrphlip]

Well, this solution (and some of the others in the comments) is way simpler than the one I came up with... mine still has the same punchline, of showing something is a factor of 3, but I took a much more roundabout way to get there...

x² + 3x + 9 = 9y²
three of these terms are multiples of 3, so the x² term must also be a multiple of 3, so x is a multiple of 3
substitute in x = 3a
9a² + 9a + 9 = 9y²
a² + a + 1 = y²

the LHS here is odd (regardless of the parity of a), so y² is odd, so y is odd
substitute in y = 2b + 1
a² + a + 1 = (2b + 1)² = 4b² + 4b + 1
a² + a = 4b² + 4b
(NB: if we stop here and add an "=0" to this equation, we quickly find our four solutions... but we still have to continue to see if there are any other solutions)

thinking: since the LHS grows like a² and the RHS grows like 4b², we should expect that a is roughly (but not exactly) equal to 2b
to explore this, substitute a = 2b + c (and we're expecting c to not be very big)
(2b + c)² + 2b + c = 4b² + 4b
4b² + 4bc + c² + 2b + c = 4b² + 4b
4bc + c² - 2b + c = 0

we've managed to eliminate our b² term, this is now linear in b, so let's solve for b:
(4c - 2)b + (c² + c) = 0
b = -(c² + c)/(4c - 2) = -c(c+1)/2(2c-1)
So now we've reduced our original diaphantine equation to a question of: when is this fraction an integer?

We can ignore the 2 in the denominator for the moment, as the numerator is always even, but look at that (2c-1) term in the denominator. In order for this fraction to be an integer, (2c-1) must be a factor of the numerator. Which means all of its prime factors must be included in the prime factors of either of the two values in the numerator, c and c+1.

However, it's not hard to show that c and 2c-1 are coprime: if any p divides c, then that p must divide 2c, so it can't divide 2c-1. Meanwhile, the gcd of c+1 and 2c-1 can be at most 3: if any p divides c+1, then it must divide 2c+2, so in order to divide 2c-1 as well, it must divide 3. So the only prime factor (2c-1) can have is at most a single 3.

So, our four potential values for 2c-1 are: -3, -1, 1, 3
We plug those values into our formulae to get values for our other variables, and check that they are, in fact, integers:
c = -1, 0, 1, 2
b = -c(c+1)/2(2c-1) = 0, 0, -1, -1
a = 2b + c = -1, 0, -1, 0
x = 3a = -3, 0, -3, 0
y = 2b + 1 = 1, 1, -1, -1

So we have the solutions: (-3, 1), (0, 1), (-3, -1), (0, -1)

[elkincampos]

First note that 3 divides x and we can suppose y>=0. Let x=3*z then
z^2+z+1=y^2;
4*z^2+4*z+4=4*y^2
(2*z+1)^2+3=4*y^2
But 3=1.3=(-1)*(-3)=3*1=(-3)*(-1)
Then we take 4 linear equations.

[benjaminvatovez]

Thank you for the video. We could have multiplied both sides by 4 and then completed the perfect square:
(2x+3)²+27=36y²
then (6y-2x-3)(6y+2x+3)=27 then there are 8 cases to check.
It could even have been shorter if we see that 3 divides x² then 3 divides x=3X thus X²+X+1=y². After multiplying by 4 as above, we get (y-2X-1)(y+2X+1)=3 which gives only 4 cases to check.

[dneary]

You can get to the answer quicker by rewriting this as (6y)^2 - (2x+3)^2 = 27 and looking at all the integer factors of 27.

[Qermaq]

Love finding the ones I missed. My approach was to look at 9y^2 as congruent to 0 mod 3. 3x and 9 are always 0 mod 3 so don't need them to solve whether x^2 can be 0 mod 3. When x^2 is congruent to 1 or 2 mod 3, its square is 1 mod 3. So only x values are mod 3. Knowing it would be a small value, tried 0, -3 and 3. Extending beyond +-3, we find that y is always going to be 2^k -1, and other than 1 and -1 we never get a square out of that. A weird way to solve it with a lot of jumps but worked for me.

[danelrosen]

I'm addicted to your videos lol, i love this channel

[SG]

Used a different approach. Tbh I didn't check if anybody else described this since I did not read all of the comments. I multiplied both sides of the equation by 4, leaving us with 4x^2+12x+36=36y^2. Now we can build a perfect square with all terms containing x leaving the remainder as a number. Means we get (2x+3)^2+27=(6y)^2. And voila, we can now get an equation with the difference of two square which we can easily factor:
(6y)^2-(2x+3)^2=27
(6y+2x+3)(6y-2x-3)=27
We now can look for factors for 27 leaving us with 8 cases including the negative factors. Bottom line we will get the same 4 solutions.

[RexxSchneider]

You can save all the waffle between 3:50 and 6:00 by simply observing that the square root of an integer is either an integer or an irrational - it is never a non-integral rational. So you set 4y² - 3 = k² immediately and save over 2 minutes. You then have x = 3(-1 ± k)/2 which requires k to be an odd integer. Then do your difference of two squares and all of the solutions fall straight out.

[SK-calc]

Nice solution.
My only criticism is that this is a very very overcomplicated solution.
There is no need for the quadratic formula.
The equation is easily factorable when multiplied by 4.

[irajnaghash]

Look when see under radical just put Y=1 then immediately find the solution X=0 or X=-3

[shanmugasundaram]

Treating y constant the given equation is a quadratic in x and by the fundamental theorem of algebra it has two solutions only.Luckily the solutions are integers.Since y is a square there are only 4 pairs of solutions.If you drop the word Diaphantic there are infinite pairs of solutions.