# Is it solvable?

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. Note by Md Mehedi Hasan
2 years, 2 months ago

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- 2 years, 2 months ago

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

Please give me an accurate solution.

- 2 years, 2 months ago

@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.

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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?

- 2 years, 2 months ago

Then we get many value of $k$

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?

- 2 years, 2 months ago

$2000$ can be expressed as the product of 2 positive integers, $k \times (m^2 + n^2)$. So the possible values of $k$ are the positive factors of 2000, namely

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

Can you carry on from here?

- 2 years, 2 months ago

I think, then with the values of $k$ I have to find the $m^2+n^2$ values and I have to continue this process and have to find $m,n$'s value too, and for every $(m,n)$ I get one solution for this equation, isn't it??

- 2 years, 2 months ago

Note that $(a,b,c) = \big( k(m^2 - n^2) , k (2mn) , k(m^2 + n^2) \big)$ or $\big( k (2mn) , k(m^2 - n^2) , k(m^2 + n^2) \big)$, with $m > n$.

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 $(a,b) = \big(4 (22^2 - 4^2), 4(2 \cdot 22 \cdot 4) \big) , \big(4 (20^2 -10^2), 4(2 \cdot 20 \cdot 10) \big) = (1872, 704), (1200, 1600)$.

- 2 years, 2 months ago

Thanks a lot. I got it. But a question, among them

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

which value I should take for $k$?

- 2 years, 2 months ago

You got to try for all of them!

- 2 years, 2 months ago

Can I escape $1$ and $2000$ for $k$? Because if I take them, then I turn back my question again

- 2 years, 2 months ago

Yes, but it's better to have tested them all just to make sure you have run through all possible combinations.

- 2 years, 2 months ago

Many many thanks to you..... I realize it. Thanks a lot.

- 2 years, 2 months ago

@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.

- 2 years, 2 months ago