Is it solvable?

Can anyone give me the solution of this question?


If a,b,ca,b,c are integer and c=2000c= 2000 and aba\ge b

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

a2+b2=c2\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
1 year, 11 months ago

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1 vote

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Hint: Parametrization of Pythagorean Triplets.

Pi Han Goh - 1 year, 11 months ago

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

Please give me an accurate solution.

Md Mehedi Hasan - 1 year, 11 months ago

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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.

The Beginner - 1 year, 11 months ago

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

Md Mehedi Hasan - 1 year, 11 months ago

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Hint: c=2000c=2000 can be expressed as k(m2+n2)k (m^2 + n^2) for positive integers k,m,nk,m,n. Now, what can the possible values of kk be?

Pi Han Goh - 1 year, 11 months ago

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@Pi Han Goh Then we get many value of kk

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

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

I read wiki there are no any solution for kk.

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

Md Mehedi Hasan - 1 year, 11 months ago

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@Md Mehedi Hasan 20002000 can be expressed as the product of 2 positive integers, k×(m2+n2)k \times (m^2 + n^2) . So the possible values of kk 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. 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?

Pi Han Goh - 1 year, 11 months ago

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@Pi Han Goh I think, then with the values of kk I have to find the m2+n2m^2+n^2 values and I have to continue this process and have to find m,nm,n's value too, and for every (m,n)(m,n) I get one solution for this equation, isn't it??

Md Mehedi Hasan - 1 year, 11 months ago

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@Md Mehedi Hasan Note that (a,b,c)=(k(m2n2),k(2mn),k(m2+n2))(a,b,c) = \big( k(m^2 - n^2) , k (2mn) , k(m^2 + n^2) \big) or (k(2mn),k(m2n2),k(m2+n2)) \big( k (2mn) , k(m^2 - n^2) , k(m^2 + n^2) \big) , with m>nm > n.

Suppose we take k=4 k =4 , then m2+n2=ck=20004=500m^2 + n^2 = \dfrac ck = \dfrac{2000}{4} = 500 .

The positive integer solutions of (m,n)(m,n) satisfying m2+n2=500m^2 + n^2 = 500 is (m,n)=(22,4),(20,10) (m,n) = (22,4) , (20, 10) .

Thus, 2 possible solutions to a2+b2=20002 a^2 + b^2 = 2000^2 are (a,b)=(4(22242),4(2224)),(4(202102),4(22010))=(1872,704),(1200,1600)(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) .

Pi Han Goh - 1 year, 11 months ago

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@Pi Han Goh 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. 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 kk?

Md Mehedi Hasan - 1 year, 11 months ago

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@Md Mehedi Hasan You got to try for all of them!

Pi Han Goh - 1 year, 11 months ago

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@Pi Han Goh Can I escape 11 and 20002000 for kk? Because if I take them, then I turn back my question again

Md Mehedi Hasan - 1 year, 11 months ago

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

Pi Han Goh - 1 year, 11 months ago

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@Pi Han Goh Many many thanks to you..... I realize it. Thanks a lot.

Md Mehedi Hasan - 1 year, 11 months ago

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@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.

Md Mehedi Hasan - 1 year, 11 months ago

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