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#1 2013-04-17 00:06:16

Agnishom
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Sum of roots

How many ordered pairs of non-negative integers (a,b) are there such that √a+√b=√432 ?


Since, √432 = 12√3, therefore the pairs are 0  + (12√3)^2, (√3)^2 + (11√3)^2, (2√3)^2 + (9√3)^2, and so on. So a total of 12 ordered pairs.

Why is this wrong?


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#2 2013-04-17 00:17:13

Nehushtan
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Re: Sum of roots

Agnishom wrote:

Since, √432 = 12√3, therefore the pairs are 0  + (12√3)^2, (√3)^2 + (11√3)^2, (2√3)^2 + (9√3)^2, and so on. So a total of 12 ordered pairs.

Wont there actually be 13 ordered pairs (because of the 0)?


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#3 2013-04-17 00:18:56

bobbym
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Re: Sum of roots

I am getting 6 with a>b.

13 with order counting.

I counted them up.

{0,432}
{3,363}
{12,300}
{27,243}
{48,192}
{75,147}
{108,108}
{147,75}
{192,48}
{243,27}
{300,12}
{363,3}
{432,0}

Oh, sorry it already got answered.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

#4 2013-04-17 00:29:01

Agnishom
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Re: Sum of roots

Oh yeah! I missed the question totally


'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
'Who are you to judge everything?' -Alokananda

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