Solving 16^{x^2+y}+16^{x+y^2}=1

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[amberdeshbhratar]

This question was asked in RMO(Regional Math Olympiad) of India .

[StaR-uwdc]

The second method (completing the square): = 1, then = 1, = 1. The first part of LHS is >=0, the exponent in the second part is >=-1/2 so the second part >=(2)4^(-1/2)=(2)(1/2)=1 and finally LHS>=1. The only case when LHS=RHS is x=-1/2 and y=-1/2 and x^2+y=x+y^2 (the last one is true when x=y=-1/2).

[titfortat]

A lot of people seem to think that symmetry of the equation implies that x must equal y for any solution (x, y). This isn't true in general. A counter-example would be the equation
(y - x - 1) * (x - y - 1) = 0
which is symmetric in x and y but holds only for x, y where either y = x + 1 or x = y + 1. Never for y = x.

[adandap]

Cool problem. I didn't think of the AM-GM trick (I often don't think to use that!) but got it this way: write 16 = 2^4, then the LHS is a sum of powers of two. The only way to get them to add to 1 is if both factors are individually 1/2. So 4(x^2 + y) = 4(y^2 + x) = -1 & eliminating y gives x^4 + 1/2 x^2 + x + 5/16 = 0. The mark 1 eyeball spotted that x = -1/2 is a root, which gives y = -1/2 (x and y are the same due to symmetry). There must be at least one more real root when you factor out (x+1/2) to get a cubic, which is also x = -1/2 as a repeated root. The remaining quadratic has no real roots.

[yoav]

Very nice!! As you mentioned, there are also nice complex sol like: x=0.5+i, y=0.5-i

[xekind]

Maybe thinking about rational numbers can provide some insights.

I think we can prove that x=y=-1/2 is the only non-complex solution by observing that 1 is rational, and that if x and y have values such that either 16^(x^2+y) or 16^(y^2+x) is irrational, it can't be a solution. (As others noted, there are also complex solutions.)

I started by substituting "p" for "x^2 + y" and "q" for "x + y^2" and trying to see what values for p and q were possible. -1/4 = p = q worked. -1/4 = p = q also implies x = y, resulting in x^2 + x = -1/4 which I solved using the quadratic formula.

But I also wanted to try lowering or raising p slightly to see what happens to q. When 0 > p > -1/4, q must be < -1/4 and vice versa, so if there is another solution besides -1/4 for p or q, one of either p or q must lie between 0 and -1/4 exclusive. 16^(-1/4) means one over the fourth root of 16 = 2. This is the highest root of 16 that yields a rational result. That means there is no number for p or q between 0 and -1/4 that yields a rational result, so -1/4 is the only value for p and q, and -1/2 the only value for x and y that yields a solution.

[ygfhfvh]

My solution is this
substitute x²+y with log_16 (a) and x+y² with log_16 (1-a)
x²+x+y²+y = log_16 (a-a²)
since a is real, a-a² ≤ 1/4
log_16 (a-a²) ≤ -1/2
x²+x+y²+y ≤ -1/2
since x and y are real, we can say that
x²+x+1/4 + y²+y+1/4 ≥ 0
x²+x+y²+y ≥ -1/2
Thus, x²+x+y²+y = -1/2
x = y = -1/2

[mathswan]

The equation remains unchanged if interchange x and y
Just solve 2*16^(x+x^2)=1
hence we have x=y=-1/2

[WahranRai]

Expression is symmetric with respect to x and y ----> x=y ----> 16^(x^2+x) = 16^(y^2+y) = 1/2 --->
Take log base 2 of LHS and RHS ---->4*(x^2+x) = -1 ----> 4x^2+ 4x + 1 = 0 ---->(2x + 1)^2 = 0 x=y= -1/2

[aoughlissouhil]

I miss integrals!!
Nice video by the way 😊!!!

[POPFIDODIDO]

1/4 = 16^(-1/2) then the whole exponent would be equal -1/2 .. 1 step simple, no log .. I was waiting the whole time you would do this

[Chester]

Solving two variables with one equation? You are a MathMagician :-D

[pedrovargas]

I smoked a joint and went that the numbers of the sum were equal, 1/2 each. 🤣

[Romik]

One more solution is to see that the equation is symmetric, which means that if you swap the variables - nothing changes. From that we get that x and y are equal. Then we substitute single variable, for example x, and get a quadratic equation for it, and solving we're getting the solution x=-1/2, which completes the solution.

[bollyfan]

Maybe try:
sin^2(theta) = 16^{x^2+y}
cos^2(theta) = 16^{x+y^2)

or basically:
sin(theta) = 4^{x^2+y}
cos(theta) = 4^{x+y^2)

Then try solving that

[patrick-]

It​ was​ equ infinity(อนันต์)​ 's​
X^3+X^2Y+XY+Y^3=16

[ahmadmazbouh]

6:35 i was commenting that when you did it

[epimaths]

Cách giải phương trình dựa vào biến đổi theo hằng đẳng thức đại số. Cảm ơn.

[mohiuddinmondal]

This is from Black Book by Vikas Gupta

[indarapusridhar]

Its question from black book functions