A Nice Diophantine Equation

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
x^2+x=2^y+1
#NumberTheoryProblems #DiophantineEquations
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

SIMILAR PROBLEMS 🤩

PLAYLISTS 🎵 :

[yoav]

Second method x^2+x=even so y=0 and x=-2, 1 😃💯

[josephsilver]

The method not mentioned is graphing the original equation. Using this method the graph quickly results in the two solutions of (1, 0) and (-2, 0), and also clearly and quickly shows these are the ONLY real solutions.

[Qermaq]

Hmm, x^2 + x is x(x+1), so if x is an integer, either x or x+1 is even. So x^2+x is even, and so 2^y+1 is also even, so 2^y is odd. For that to happen with y being an integer, y has to be 0. 2^0+1=2.

x^2 + x - 2 = 0 is a simple quadratic where (x-1)(x+2)=0, , so x = 1 or -2.

[marklevin]

X^2+x is always an even number. Therefore 2^y+1 must be even too. This can be only when y=0. If y is negative then 2^y is not an integer and if y is positive then 2^y must be even and 2^y+1 is odd.
Assuming y =0. we are getting quadratic equation for x. Roots are x=1 or x =-2.

[seanfraser]

x^2 + x is always even. This is because x^2 + x = x(x+1). Either x or x+1 is even, and the product of an even integer with any other is always even.

If y > 0, 2^y is even and thus 2^y + 1 is odd. If y < 0 then 2^y + 1 is not an integer. So y = 0. We thus need to find x such that x^2 + x = 2. By any method you want, you find x=1 and x=-2 are the solutions to that quadratic.

So the only two integer solutions are (1, 0) and (-2, 0)

[dwm]

Quickly we get that y = 0, only [for reasons of odd & even discussed in video]
If y = 0, RHS = 2
Therefore, x^2 + x - 2 = 0, a quadratic with two integral solutions
so, (x - 1)(x + 2) = 0 and x = 1 or x = -2

[damiennortier]

x = (-1 +- sqrt(1 + 4×2^y + 4))/2 = (-1 +- sqrt(5 + 4×2^y))/2
Since we're looking for integer, 5 + 4×2^y must be a perfect square. Specifically, 4×2^y cannot be irrationnal. So y need to be greader than -2. By checking 1, 2, 0, -1, -2, we find that y = 0 work. How to prove that there are no other solution or are there any other solutions ?
First, consider that y is even like 2n.
After substituding, we find that 5 + 4×2^y = 5 + 4×(2^n)^2. The only square that give you another square when we add 5 is 4. So n = 0 give you y = 0.
Second case, consider that y is odd like 2n + 1. Then, 5 + 4×2^y = 5 + 6×(2^n)^2 = (3 + sqrt(6)×2^n)^2 - 6sqrt(6)×2^n
But I don't know if this give us any integer solution.
So y = 0 gives us x = 1 or x = -2

[marcushendriksen]

At a glance, if x and y are integers, then the LHS must always be even. The only way that 2^y + 1 can ever be even is if 2^y itself is odd, i.e. y = 0. Thus we have x^2+x = 2 and there are only two solutions to that equation: x = -2 or x = 1. (-2, 0) and (1, 0) is the answer.

[yoav]

For the first method : for n>=3 2^n+5=5(mod8) but m^2=0, 1, 4 (mod8) so there are no other solutions.

[BlaqRaq]

I was expecting a whole number of pi to show up since we were looking from different angles. 😂😂😂😂

[damonpalovaara]

I isolated the 2^y then took the log of both sides. That gave me y = log_2(x^2 +x -1). I realized that x^2 + x -1 is always odd and the only odd number that has an integer log_2 value was 1 which gives a y value of 0. So now you have to solve x^2 + x - 1 = 1 which has solutions x=-2 and x=1. Therefore the solutions are (-2, 0) and (1, 0)

[jisuuqm]

x(x+1)=2 => x=1 or x=-2 so (x, y)=(-2, 0) is also another solution.

[goldfing]

We can rewrite the left side:
x*(x + 1) = 2^y + 1
For y = 0, we get x^2 + x = 2^0 + 1 = 1 + 1 = 2.
Clearly, x = 1 fulfills this, because 1^2 + 1 = 1 + 1 = 2.

But so does x = -2, because (-2)^2 + (-2) = 4 - 2 = 2.
The equation
x^2 + x = 2 or
x^2 + x - 2 = 0 or
(x - 1)(x + 2) = 0
cannot have other solutions than these two.
So (x = 1, y = 0) and (x = -2, y = 0) are solutions.


However, x*(x + 1) on the left side is always even, so there cannot be any integer solution for x.

For y < 0, 2^y is no integer, thus 2^y + 1 is no integer.
However, the left side x*(x + 1) is always integer, thus no solution for x.

I think there are only the two solutions mentioned above.

[devondevon]

y=0 and x=1
x^2 + x for all integers will be positive
2^ y will be positive except when y=0; hence 2^y + 1 is odd
x^2 + x = 2^y +1
x^2+ x -2 = 2^y-1
let x =1
1 + 1 -2 =0
0 = 2^y-1
1 = 2^y
0 = y

[nicolascamargo]

x^2+x=(2^y)+1 es equivalente a x(x+1)=(2^y)+1, ya que el lado izquierdo es producto de consecutivos, si x es par da par por impar que es par y si x es impar da impar por par que es par de donde el lado derecho también debe ser par y la única forma que suceda esto con números enteros es con y=0 ya que con y≥1, (2^y)+1 es impar y con y≤-1, (2^y)+1 se cumple la desigualdad 1<(2^y)+1<2. Lo que falta es reemplzar este valor y=0 en la ecuación quedando x^2+x=2 que sumando (1/4) en ambos lados y expresando el (9/4) del lado derecho como (3/2)^2 queda: (x+(1/2))^2=(3/2) de donde x=-(1/2)±(3/2) quedando x=1 o x=-2 como el enunciado son soluciones enteras entonces estás son dos (-2, 0) y (1, 0).

[bkkboy-cmeb]

at y=0 → x=1, -2
at y≠0 → x²+x ≡ 1(mod2) → no solution.

[JeremyLionell]

Can anyone explain what Diophantine mean?

[trnfncb]

LHS is x(x+1) so it is even. RHS is odd unless y=0

[goldfing]

What about x = -2, y = 0 ? Does Diophantine mean "only integer numbers" or "only natural numbers"?

[ghstmn]

Cant y be another number tho? And if not, shouldnt we prove it?