Only 1 percent can solve this

Vedantu Math video
Math Stack Exchange

Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!

If you purchase through these links, I may be compensated for purchases made on Amazon. As an Amazon Associate I earn from qualifying purchases. This does not affect the price you pay.

Book ratings are from January 2022.

My Books (worldwide links)

My Books (US links)
Mind Your Decisions: Five Book Compilation
A collection of 5 books:
"The Joy of Game Theory" rated 4.2/5 stars on 224 reviews
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 4/5 stars on 24 reviews
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4.1/5 stars on 38 reviews
"The Best Mental Math Tricks" rated 4.2/5 stars on 76 reviews
"Multiply Numbers By Drawing Lines" rated 4.3/5 stars on 30 reviews

Mind Your Puzzles: Collection Of Volumes 1 To 3
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 87 reviews
"Math Puzzles Volume 2" rated 4.1/5 stars on 24 reviews
"Math Puzzles Volume 3" rated 4.2/5 stars on 22 reviews

2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.

My Blog

Twitter

Merch

Patreon

Press

[mrmango]

Excellent problem.
I solved using another method which is fairly elegant I would say :), though not as much as yours. It involves reduction to the parametric equation of a circle. No AM, GM inequality.

Rewrote the equation as (4^(x^2+y))^2 + (4^(x+y^2))^2 =1
This is the equation of a circle A^2+B^2=1 in the variables A= 4^(x^2+y), B= 4^(x+y^2)
Now I used the circle parametric equation in the parameter angle “t”.

4^(x^2+y) = sin(t)
4^(x+y^2) = cos(t)

Observe that the LHS in both the equations is greater than zero, because 4^(some real number) > 0.
So the RHS terms sin(t), cos(t) are both strictly positive.
Apply log (to base 2 everywhere) to both equations. (We can do that because the terms are>0)

x^2 + y = (1/2) log (sin(t))
x + y^2 = (1/2) log (cos(t))

Add both of the equations.

x^2 + x + y^2 + y = (1/2) log (sin(t)cos(t))

Use the expansion of sin (2t) = 2 sin(t) cos(t)

x^2 + x + y^2 + y = (1/2) log(sin2t) - (1/2)
<= -1/2

because sin(2t) <= 1, log(sin2t) <= 0
(<= is the sign for less than equal to) Now, we have

x^2 + x + y^2 + y <= -1/2
x^2 + x + y^2 + y + 1/2 <= 0
(x+0.5)^2 + (y+0.5)^2 <=0

Since x, y are real the only solution is x=-0.5, y=-0.5

[titan]

Presh, my hat goes off to you because you're humble enough to admit, as a professional mathematician, that there are problems that you can't solve. Me neither, I couldn't solve this problem, especially come up with such a brilliant solution as using an inequality. Thank you!

[deerho]

It's fascinating that to solve an equation, you start with an inequality.

[tilakrupareliya]

When I saw the thumbnail, I immediately thought for "x=y" because of the symmetry of the equation.
After putting x=y, you get x = y = -1/2 = -0.5
But I was not sure that it is the only solution.
Good video 👍

[poomonyoutube]

This is so beautiful. Simplifications like this is why I love algebra, the way something so complicated turns out to be something so simple... It's mesmerizing

[allisthemoist]

These are so good for keeping the brain good at solving problems.

[rajinisekar]

I got the correct answer but my method of getting there is definitely wrong. I used this method when I figured I couldn't find 'all' real solutions, so decided to at least get one simple one

16^(x^2 + y) + 16^(x + y^2) = 1
If both terms were equal to 0.5, you'd have at least one of the possible answers, so, because 16 to the power of both terms is equal to 0.5:
x^2 + y = -0.25
y^2 + x = -0.25
Rewrite them both to:
x^2 + y + 0.25 = 0
y^2 + x + 0.25 = 0
putting together gives you
x^2 + y + 0.25 = y^2 + x + 0.25
Which gives x = y
Substituting x into y:
x^2 + x + 0.25 = 0
That gives (x + 0.5)^2 = 0,
So x = -0.5 and y = -0.5

Turns out to be the only right answer, but if I got this in my actual tests and there were more answers, I'd be fucked 😅

[trnfncb]

The expression is symmetric about x=y. So let x=u+d and y=u-d with d>0. You can immediately rewrite 16^(u^2+u+d^2)[16^a+16^(-a)], with a=d(2u-1). The expression in square brackets (like any z^t+z^(-t)) is >=2, while the factor outside has a minimum of 16^(-1/4)16^(d^2) = 16^(d^2)/2 for u=-1/2. So the expression is always >=1, with equality holding only if d=0, i.e. x=y=u=-1/2.

[unknownredacted]

It never once occurred to me to use the A.M>/G.M>/H.M principle to solve this question

[HamishArb]

I couldn't prove it (although I didn't try for very long), but I did find the solution like so in about a minute.
Notice that x and y are swappable (which is probably not the technical term lol, here meaning that if you swap x and y in the equation you get the same equation back, which also means that if you have x=1, y=2 as a solution for example, you immediately get x=2, y=1), therefore x=y is likely to be a solution, I checked if setting x=y makes sense and it was easy to find.
Obviously this method doesn't guarantee every solution to be found nor does it guarantee any solutions at all, but it works quite often for these sorts of problems.

[anuragmudgal]

thats a Regional Mathematics Olympiad(India) 2011 problem btw. Problem 6. pretty cool.

[paparmar]

So this is a very symmetric problem, as others have pointed out. Since the roles of x and y are interchangeable, we know that (x^2 + y) and (y^2 + x) are the same (note this does not necessarily mean x =y). This also implies that both LHS terms are equal (i.e., to 0.5 = 16^-0.25). At this point, we can infer all solutions to x and y must be negative fractions. Using 16 = 2^4, we then get the equations of two parabolas : one open-to-the-bottom vertical (y = -x^2 - 0.25) and one open-to-the-left horizontal (x = -y^2 - 0.25). Note the symmetry between x & y remains as it must. Now we can infer that there are at most four solutions, since such parabolas can intersect at zero, 1, 2, 3 or 4 points. It turns out these two parabolas have exactly one intersection, namely x = -0.5, y = -0.5. You can get the final solution by algebraic manipulations as detailed in this video, or by plotting (I used wolframalpha - but I wouldn't do that until I had distilled the original problem to this level). Personally, I like to focus on the properties of the potential solutions as much as I can before getting to their specific values.

[NestorAbad]

What an elegant solution! Thanks for sharing.

I used a different approach to solve it, not so elegant but it worked:

First, let (x, y) be a solution of the equation 16^(x^2+y)+16^(x+y^2)=1, then I look for the region in the plane where this solution must lie. To do this, note that both exponents x^2+y and x+y^2 must be negative. This is because 16^z>0 for all real z, and if z>=0, then 16^z>=1. So if one of the exponents was nonnegative, then 16^(x^2+y)+16^(x+y^2) >= 1 + positive > 1.

So, the inequalities x^2+y<0 and x+y^2<0 give us a beautifully symmetrical blade-shape little region contained in quadrant III, specifically inside the square (-1, 0)×(-1, 0).

The easiest point that we can check inside this region, and for the sake of symmetry, is (-1/2, -1/2). If we toss these coordinates into 16^(x^2+y)+16^(x+y^2) we get 16^(1/4-1/2)+16^(-1/2+1/4) = 2·16^(-1/4) = 2·1/2 = 1, bingo! (We could also have obtained this solution by forcing 16^(x^2+y)=16^(x+y^2)=1/2, so solving x^2+y=x+y^2=-1/4.)

Now our goal is proving that this solution is unique. Let's consider (x, y)=(-1/2+a, -1/2+b) where a, b are any real numbers, and we'll see that this point, when substituted into 16^(x^2+y)+16^(x+y^2), gives us something >1 except when a=b=0:

16^((-1/2+a)^2-1/2+b) + 16^(-1/2+a+(-1/2+b)^2) =
= 16^(1/4-a+a^2-1/2+b) + 16^(-1/2+a+1/4-b+b^2) =
= 16^(-1/4-a+a^2+b) + 16^(-1/4+a-b+b^2) =
= 16^(-1/4)·(16^(-a+b+a^2) + 16^(a-b+b^2)) =
= 1/2·(16^(-a+b+a^2) + 16^(a-b+b^2)) >=
>= 1/2·(16^(-a+b) + 16^(a-b)) =
= 1/2·(u + 1/u)

where we made the substitution u=16^(-a+b). Now note that u>0, and it's easy to see that for u>0 the real function f(u) = u + 1/u reaches its absolute minimum at u=1, f(u)=2, so we can end our chain of operations writing

1/2·(u + 1/u) >= 1/2·2 = 1

and the >= is an = when u=1, that is, when 16^(-a+b)=1, so -a+b=0 and a=b. But in that case, if we recall our last expression before the first >= and put a=b we get:

1/2·(16^(-a+b+a^2) + 16^(a-b+b^2)) = 1/2·(16^(a^2) + 16^(a^2)) = 16^(a^2),

and if we want 16^(a^2)=1 implies a=0, so a=b=0, as we wanted to show.

[venkatramana]

As a math and physics faculty.... I appreciate you bless you..

[rjmunro]

It would have been good to put the numbers from the end in to the equation at the beginning to prove it works:
x^2 + y = x + y^2 = (1/4) + (-1/2) = -1/4
16 ^ (-1/4) = 1/2
(1/2) + (1/2) = 1

[SyberMath]

Interesting! I made a video on this problem back in April 2022. I don't think it was a suggestion. I must have seen it in a book somewhere. Anyways, I think the way you write 16 as 4^2 at 2:27 and the rest of it with 0.5 being powers seems a little confusing for the average viewer.
Overall a good video!

[WoodyC-fvhz]

(-0.5)=x=y. Since nothing suggests that x and y cannot be of the same value and the terms on either side of the '+ sign' can't be equal, I looked for a single exponent for 16 achieving the final value of 0.5 for both terms. This can be split into (-0.5) + 0.25 for both exponents, resulting in 16^(-0.25) + 16^(-0.25) = 1.

[methodiconion]

I did this differently. I assumed x=y (figured while it may not necessarily be true, there is a solution such that it is.). 16^(x^2+x)=0.5 => x=-1/2 +-

[Fred-yqfs]

A tough one!
I quickly found it'd be nice to deal with x2+y2+x+y as it factors nicely as a circle expression (x+0.5)^2+(y+0.5)^2 - 1/2, but I could not find how to get to that.
I also knew I'd like to tackle this with an inequality as -1/2 was obviously a solution sitting there as an extremum with x and y constrained between the y=-x^2 and y = +-sqrt(-x) for x and y between -1 and 0.
I said to myself: how to go from this sum to a product... but never did think about the am/gm inequality! Bummer.
You really have to have a level of mastery to think about it as a bridge between sum and product.
Thanks for that one!

[daqingli]

By watching the solution, I think the problem can also be solved another way. It is clear that in the original equation x, y are symmetrical with each other so if x = a is a solution for the equation, y=a is also a solution or there should be x = y for the solution of the equation. Thus, the equation can be written as 2(16^(x^2+x)) = 1 or 16^(x^2+x) = 1/2. Taking a log with a base of 2 to both sides: we have (x^2 + x) * 4 = -1 or x^2+x = -1/4 and can be solved from (x + 0.5)^2 = 0 thus x = -0.5 and there is also y = -0.5 as stated before.