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Oh yeah! I missed the question totally

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.

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.

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)?

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?