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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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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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 96,647

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.

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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'

'But our love is like the wind. I can't see it but I can feel it.' -A Walk to remember

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