Can anyone give me the solution of this question?

If \(a,b,c\) are integer and \(c= 2000\) and \(a\ge b\)

How many solution we get for \((a,b)\) from this equation?

\[\large a^2+b^2=c^2\]

\[\]

-If anyone want to modify the question for a better solution, I approve you to do so. I am only looking for the solution process.

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## Comments

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TopNewest@Calvin Lin sir, I read that wiki but I can't be able to apply that in this case. Please sir, give me more hints to solve it.

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Hint:Parametrization of Pythagorean Triplets.Log in to reply

@Pi Han Goh I update my post. Now can you help me, please?

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Hint:\(c=2000\) can be expressed as \(k (m^2 + n^2) \) for positive integers \(k,m,n\). Now, what can the possible values of \(k\) be?Log in to reply

Suppose, if I take \(m=2\) and \(n=4\) then we get \(k=100\)

if I take \(m=2\) and \(n=6\) then we get \(k=50\)

I read wiki there are no any solution for \(k\).

So there I have a little confusion. How can I go forward afterthat?

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\[ 1 , 2 , 4 , 5 , 8 , 10 ,16 ,20 ,25 ,40 , 50 , 80 , 100 , 125 , 200 , 250, 400 , 500 ,1000 , 2000 . \]

Can you carry on from here?

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Suppose we take \( k =4 \), then \(m^2 + n^2 = \dfrac ck = \dfrac{2000}{4} = 500 \).

The positive integer solutions of \((m,n) \) satisfying \(m^2 + n^2 = 500 \) is \( (m,n) = (22,4) , (20, 10) \).

Thus, 2 possible solutions to \( a^2 + b^2 = 2000^2 \) are \((a,b) = \big(4 (22^2 - 4^2), 4(2 \cdot 22 \cdot 4) \big) , \big(4 (20^2 -10^2), 4(2 \cdot 20 \cdot 10) \big) = (1872, 704), (1200, 1600) \).

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\[ 1 , 2 , 4 , 5 , 8 , 10 ,16 ,20 ,25 ,40 , 50 , 80 , 100 , 125 , 200 , 250, 400 , 500 ,1000 , 2000 . \]

which value I should take for \(k\)?

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@Pi Han Goh that would result in a lot of cases.. we need to consider at least 2000 values of c. That would result in lot of time... Can you please suggest any shorter method or how to proceed further with this.

Thanks in advance.

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Thanks. I read it, But I didn't clear...

Please give me an accurate solution.

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