A Diophantine Equation @pkqualitymath1234

In this video, I solved a Diophantine equation without any special tricks, just brute force.
Watch @pkqualitymath1234

[jamesharmon]

When you said "brute force", I expected you to plug integers into the given equation to see what happens.

[keinKlarname]

Your hand writing is really sensational!

[courbe]

I like your way of explaining, a real mathematics teacher, thank you.

[nanamacapagal]

BONUS: I allowed sqrt(-N) = i * sqrt(N), and that allowed for another solution (and only one more): (-2, 3)

[futuregenerationinstitute]

Thanks for making maths easier. Would you please make a video about parallelogram formulas practically using DIAGONALS. Thank you sir.

[Abby-hisf]

@12:05 I love your explanation, it gives real depth how to see possible solutions,
with brutal force
x^2 (4 - y) = 4 = (4)(1)
x^2 =4, x= 2 and 4 - y = 1, so y = 3 is

[Alephŋull]

You can also do one thing
Take 4 on rhs
Then you will have
X * X * (4-Y) = 4
One two options left
1*1*4
Or
2*2*1
1st case does not work
Therefore x=2 and y=4-1=3

[geekandnerd]

Man, i love his videos so much. Is just so great to learn from someone who really likes what he is talking about. I got into maths because of this guy when I was like 15 and have absolutely no regrets.

I don’t, know how often do you read your comments, but if you’re reading this, man, thank you. I’m currently battling to get a bachelor on maths, and I have no idea how my life would’ve turned out if I haven’t found that video of yours three years ago.

[chaosredefined]

From 2sqrt(x^2 - y) = xy - 2x.

Note that x | RHS. Therefore, x | LHS. So, x | 2 sqrt(x^2 - y). Therefore, x^2 | 4x^2 - 4y. So, x^2 | 4y. Therefore, there exist some k such that 4y = k x^2. But we know that y >= 0, so 4y and x^2 are both positive, and therefore k has to be positive.

Going back to the equation, we can move the 2 into the sqrt by squaring it, giving us sqrt(4x^2 - 4y) = xy - 2x. Then, replacing that 4y with k x^2, we get sqrt(4x^2 - kx^2) = xy - 2x. Extracting out the x^2 from the sqrt, we get x sqrt(4 - k) = xy - 2x. Option 1 is that x = 0, which leads us to 4y = k 0^2, and therefore 4y = 0, so y = 0. Option 2 is that sqrt(4 - k) = y - 2. Since k has to be positive, and sqrt(4 - k) has to be an integer, this means that sqrt(4-k) = 0, 1 or 2, or k = 4, 3 or 0.

Case 1: k = 4
4y = 4 x^2, so y = x^2. Therefore, 2 sqrt(0) = x(4x^2) - 2x, so 4x^3 - 2x = 0, or 2x(2x^2 - 1) = 0. 2x^2 - 1 has no integer solutions, so the only solution is the (0, 0) solution we already covered.

Case 2: k = 3
4y = 3x^2. So, y = 3x^2 / 4. 2 sqrt(x^2 - 3x^2 / 4) = 3x^3 / 4 - 2x => 2 sqrt(x^2 / 4) = 3 x^3 / 4 - 2x => 2 (x/2) = 3x^3 / 4 - 2x => 0 = 3x^3 / 4 - 3x => 3x( x^2 / 4 - 1) = 0. x = 0 is a solution again, which gives us y = 0. But x^2 / 4 - 1 has integer solutions, specifically x = 2 and x = -2. We reject x=-2 because we need x >= sqrt(y) >= 0. Therefore, x = 2 is the only new value we get from this case, and it gives us that y = 3 (2)^2 / 4 = 3. so, (2, 3).

Case 3: k = 0
This means that 4y = 0 x^2, so y = 0. Shoving that in, we get 2 sqrt(x^2) = -2x. But x >= 0, and now we need -2x <= 0, which gives us the same (0, 0) solution again.

So, we end up with two solutions: (0, 0) and (2, 3).

[martinmolander]

I really like your videos, they are fantastic. However...

I enjoyed this until the end; but when you have a radikal equation, you must always, ALWAYS, check your solutions to the original equation.

[assiya]

شكرا أستاذ
في الدقيقة 12 من بين الحلول أرى (1،0) لماذا لم تأخذها بعين الاعتبار وشكرا

[MrJasbur]

If it’s a Diophantine equation, isn’t the only requirement that x and y be integers? Like with the more complicated expressions involving square roots being whatever they want whether irrational or even complex? If so, wouldn’t (-2, 3) also be a solution?

[iqtrainer]

You did a fantastic job PN! You guys should do more collabs

[JPErrico]

I'm sure someone has noted to you that if x^2 = 4/(4-y), then you can also have (1, 0), but that solution doesn't work in the original equation.

[Rishab_Sharma_Python_Teacher]

Sir please make a video on how to find intersection coordinates of two circles

[janimed]

Bravo. Mais Pour la dernière réponse il fallait mieux expliquer
D'après la dernière égalité on a
4-y doit être supérieur a
D'autre part :y>=0
On déduit que y =0 ou y=1 ou y=2 ou y=3
Y=0 déjà une solution. Pour y=1et2 impossible
Y=3 une solution

[TheMathManProfundities]

Why can't you take the root of negative numbers? √-6=i√6. x=-2 is therefore a valid candidate solution and turns out to be valid (y=3). Also in this vein, you should have added a ± when you multiplied the roots together (both can be imaginary). As it happens, the next thing you did was to square this so it wouldn't have made any difference. Also, later on you divided by y-4 without checking whether this can be zero. It can easily be seen that it can't be but this should have been included especially as this is designed to teach other people.

[dawkinsfan]

03:37 😂😂😂😂😂😂 I watched it at 2x...it is even more hilarious than the original one! 😂

[JamesWanders]

For 4x^2-4-x^2•y=0, (x, y)=(1, 0) looks to be a valid solution--but that doesn't work in the original equation. I'm struggling to see why the (2, 3) solution works and (1, 0) does not as both satisfy that side of the zero product step 🤨

[dirklutz]

An "illegal", but nice solution is: (x=sqrt(2), y=2)