solve an equation

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chrislav

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solve the following equation
[math] 16^{x^2+y}+16^{x+y^2} =1[/math]
 
chrislav said:
i know one solution but i am itersting in others
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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 do it but lets wait maybe others have other ideas
yes my approach is thru AM-GM Inequality
 
No one is going to supply you with an answer until you show your solution.
 
BigBeachBanana said:
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.
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I will use AM-GM
Hence:
[math] 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[/math]Hence x=-(1/2) and y=-(1/2)
 
chrislav said:
I will use AM-GM
Hence:
[math] 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[/math]Hence x=-(1/2) and y=-(1/2)
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I don't follow how you go from
[imath]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}}[/imath]
What property or theorem did you use?
 
BigBeachBanana said:
I don't follow how you go from
[imath]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}}[/imath]
What property or theorem did you use?
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you suggested an AM-GM inequality ,
Then [math]x^2+y+x+y^2[/math]=

[math](x+\frac{1}{2})^2+(y+\frac{1}{2})^2-\frac{1}{2}[/math]
 
chrislav said:
[math] 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}}[/math]
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I believe your inequality is going the wrong way.
[math]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} [/math]This makes the rest of the argument invalid.
 
BigBeachBanana said:
I believe your inequality is going the wrong way.
[math]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} [/math]This makes the rest of the argument invalid.
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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 [math]16^{x^2+y} +16^{x+y^2}[/math] with 1 you will get my inequality
3rd in your solution you made an assumption which is based nowhere
 
chrislav said:
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 [math]16^{x^2+y} +16^{x+y^2}[/math] with 1 you will get my inequality
3rd in your solution you made an assumption which is based nowhere
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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.
[math]\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[/math]
 
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BigBeachBanana said:
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.
[math]\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[/math]
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I reapet again if you take
[math]\frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}[/math]And substitute [math] 16^{x^2+y}+16^{x+y^2}[/math] with 1 what do you get?
if you have an inequality [math]A\geq B[/math] AND you substitute A WITH C dont you get :[math]C\geq B[/math]?
 
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chrislav said:
I reapet again if you take
[math]\frac{16^{x^2+y}+16^{y^2+x}}{2}\ge \sqrt{16^{x^2+y}\times16^{y^2+x}}[/math]And substitute [math] 16^{x^2+y}+16^{x+y^2}[/math] with 1 what do you get?
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No need to repeat. I got it the first time you said it. Also, I'm done here.
 
Is there anybody else who finds my solution wrong
Please be so kind as to point out where is the mistake
 
chrislav said:
Is there anybody else who finds my solution wrong
Please be so kind as to point out where is the mistake
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Your problem is that you are leaving out too many details of your proof for people to follow. See here for the details.

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

-Dan
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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
 
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