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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: 81,656

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'

'Humanity is still kept intact. It remains within.' -Alokananda

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