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#1 2013-04-16 02:06:16

Agnishom
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From: The Complex Plane
Registered: 2011-01-29
Posts: 16,229
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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 2013-04-16 02:17:13

Nehushtan
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From: London
Registered: 2013-03-09
Posts: 612
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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.

Won’t there actually be 13 ordered pairs (because of the 0)?


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#3 2013-04-16 02:18:56

bobbym
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From: Bumpkinland
Registered: 2009-04-12
Posts: 85,270

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.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#4 2013-04-16 02:29:01

Agnishom
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From: The Complex Plane
Registered: 2011-01-29
Posts: 16,229
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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'
'Humanity is still kept intact. It remains within.' -Alokananda

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