solve an equation

[chrislav]

solve the following equation
$$16^{x^2+y}+16^{x+y^2} =1$$

 

[BigBeachBanana]

Is this a challenge or do you need help solving it?

 

[chrislav]

i know one solution but i am itersting in others

 

[BigBeachBanana]

i know one solution but i am itersting in others

By "solution" do you mean you found a pair (x,y) that satisfy to the equation, but you're wondering if there's more? Or do you mean you want to see an approach that is different from yours? In either case, please share your solution.

PS: I see an AM-GM inequality.

 

[chrislav]

i will do it but lets wait maybe others have other ideas
yes my approach is thru AM-GM Inequality

 

[Steven G]

No one is going to supply you with an answer until you show your solution.

 

[topsquark]

i know one solution but i am itersting in others

There is only one real solution for (x, y). But there are many complex solutions. Are these what you are trying to find or are you trying to prove there is only one real solution?

-Dan

 

[chrislav]

By "solution" do you mean you found a pair (x,y) that satisfy to the equation, but you're wondering if there's more? Or do you mean you want to see an approach that is different from yours? In either case, please share your solution.

PS: I see an AM-GM inequality.

I will use AM-GM
Hence:
$$16^{x^2+y} +16^{x+y^2}=1\implies \frac{1}{4}\geq 16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}}\implies16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2}\leq 1\implies (x+\frac{1}{2})^2+(y+\frac{1}{2})^2\leq 0$$Hence x=-(1/2) and y=-(1/2)

 

[BigBeachBanana]

I will use AM-GM
Hence:
$$16^{x^2+y} +16^{x+y^2}=1\implies \frac{1}{4}\geq 16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}}\implies16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2}\leq 1\implies (x+\frac{1}{2})^2+(y+\frac{1}{2})^2\leq 0$$Hence x=-(1/2) and y=-(1/2)

I don't follow how you go from
$16^{x^2+y} +16^{x+y^2}=1\implies \frac{1}{4}\geq 16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}}$
What property or theorem did you use?

 

[chrislav]

I don't follow how you go from
$16^{x^2+y} +16^{x+y^2}=1\implies \frac{1}{4}\geq 16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}}$
What property or theorem did you use?

you suggested an AM-GM inequality ,
Then $$x^2+y+x+y^2$$=

$$(x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}$$

 

[BigBeachBanana]

$$16^{x^2+y} +16^{x+y^2}=1\implies \frac{1}{4}\red{\geq} 16^{(x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}}$$

I believe your inequality is going the wrong way.
$$A.M \ge G.M\\ \frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}\\ 16^{x^2+y}+16^{y^2+x} \ge 2\sqrt{16^{x^2+y+y^2+x}}\ge 1\\ \implies \sqrt{16^{(x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 - \frac{1}{2}}}\ge \frac{1}{2}\\ \implies 16^{(x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 - \frac{1}{2}} \ge \frac{1}{4}$$This makes the rest of the argument invalid.

 

[chrislav]

I believe your inequality is going the wrong way.
$$A.M \ge G.M\\ \frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}\\ 16^{x^2+y}+16^{y^2+x} \ge 2\sqrt{16^{x^2+y+y^2+x}}\ge 1\\ \implies \sqrt{16^{(x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 - \frac{1}{2}}}\ge \frac{1}{2}\\ \implies 16^{(x + \frac{1}{2})^2 + (y + \frac{1}{2})^2 - \frac{1}{2}} \ge \frac{1}{4}$$This makes the rest of the argument invalid.

first : you should check up my solution and see if the values of x and y i found are correct ,i mean if they justify the original equation
2nd in your 1st inequality if you substitute $$16^{x^2+y} +16^{x+y^2}$$ with 1 you will get my inequality
3rd in your solution you made an assumption which is based nowhere

 

[BigBeachBanana]

first : you should check up my solution and see if the values of x and y i found are correct ,i mean if they justify the original equation
2nd in your 1st inequality if you substitute $$16^{x^2+y} +16^{x+y^2}$$ with 1 you will get my inequality
3rd in your solution you made an assumption which is based nowhere

First, I'm not questioning your answer, but rather how you got your answer.
Second, that's why I asked for your explanation in post #9.
Third, it's not based out of nowhere. It's just that you can't see it.
$$\frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}\\ 16^{x^2+y}+16^{y^2+x} \ge 2\sqrt{16^{x^2+y+y^2+x}}= 2\cdot 4^{x^2+y+y^2+x}\\ \text{Since } x^2+x \ge -\frac{1}{4}, \text{similarly for } y \text{ it follows:}\\ 2\cdot 4^{x^2+y+y^2+x}\ge 2\cdot 4^{-\frac{1}{2}}=1\\ \implies 16^{x^2+y}+16^{y^2+x} \ge 2\sqrt{16^{x^2+y+y^2+x}}\ge 1$$

 

[chrislav]

First, I'm not questioning your answer, but rather how you got your answer.
Second, that's why I asked for your explanation in post #9.
Third, it's not based out of nowhere. It's just that you can't see it.
$$\frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}\\ 16^{x^2+y}+16^{y^2+x} \ge 2\sqrt{16^{x^2+y+y^2+x}}= 2\cdot 4^{x^2+y+y^2+x}\\ \text{Since } x^2+x \ge -\frac{1}{4}, \text{similarly for } y \text{ it follows:}\\ 2\cdot 4^{x^2+y+y^2+x}\ge 2\cdot 4^{-\frac{1}{2}}=1\\ \implies 16^{x^2+y}+16^{y^2+x} \ge 2\sqrt{16^{x^2+y+y^2+x}}\ge 1$$

I reapet again if you take
$$\frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}$$And substitute $$16^{x^2+y}+16^{x+y^2}$$ with 1 what do you get?
if you have an inequality $$A\geq B$$ AND you substitute A WITH C dont you get :$$C\geq B$$?

 

[BigBeachBanana]

I reapet again if you take
$$\frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}$$And substitute $$16^{x^2+y}+16^{x+y^2}$$ with 1 what do you get?

No need to repeat. I got it the first time you said it. Also, I'm done here.

 

[chrislav]

Is there anybody else who finds my solution wrong
Please be so kind as to point out where is the mistake

 

[topsquark]

Is there anybody else who finds my solution wrong
Please be so kind as to point out where is the mistake

Your problem is that you are leaving out too many details of your proof for people to follow. See here for the details.

-Dan

 

[chrislav]

Your problem is that you are leaving out too many details of your proof for people to follow. See here for the details.

-Dan

Thank you thank you indeed for showing me a proof that is exactly the same with mine .
Now w.r.t the details of the proof, when i was asked by BigBeachBanana to show more details in his post #9
he called my inequality resulting from the use of AM-GM Inequality wrong AND hence my whole argument invalid in his post #11
Now what do you suggest i should do