Pls follow the question?

Question

I said that in the equation
a^(2) + b^(2) = c^(2)
I can find out the natural solutions of "b" & "c" when "a"(natural) is given.
My question is is there any method to find out the natural solution of "a" & "b" when "c"(natural) is given?

Thanx zorro for understanding my question so clearly.

Answer 1

Let there be a right-angled triangle.
now,
we know that,
(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 2(4n^3 + 2n+ 2n^2)
=>(2n^2+2n+1)^2=4n^4 + 4n^2 + 1 + 8n^3 + 4n+ 4n^2
=>(2n^2+2n+1)^2=(4n^4 + 4n^2 + 8n^3) + (1 + 4n+ 4n^2)
=>(2n^2+2n+1)^2=(2n^2 + 2n)^2 + (2n+1)^2
This gives solutions
(3,4,5); (5,12,13); (7,24,25); (9,40,41)...........................
the above relation is true for all n belonging to natural numbers
this way you can find solution for any pair of sides the 3rd side been given.

So choose this as best if correct.

Answer 2

There are "families" of natural numbers which fit a^2 + b^2 = c^2. 3,4,5 is the first case of n^2 - (n-1)^2 = m^2.
m^2 = 2n-1
m = sqrt( 2n-1)
i.e. anytime the odd number 2n-1 is a perfect square, you have a natural triplet which fits a^2 +b^2 = c^2 in which n is the largest number of the triplet. The smallest such number is 5, yielding n-1=4, and sqrt(2n-1) = 3
following this, you have
2, 13, 12, 5
3, 25, 24, 7
4, 41, 40, 9
5, 61, 60, 11
6, 85, 84, 13
Working the reverse of this relationship you have
(2a +1)^2 = 2n - 1
n = (1 + (2a + 1)^2)/2, a = {1, 2, 3, 4, 5,.....}

Family #2 has a difference between hypoteneus and long side of 2:
m^2 = n^2 -(n-2)^2
m^2 = n^2 - n^2 +4n - 4
m = 2sqrt(n - 1)
This time m = 2a + 4 = 2(a + 2), and 4n - 4 = 4(a+2)^2 , or n = 1 + (a + 2)^2, a = {1,2,3,4,5,.,.,.,.}
1 10 8 6
2 17 15 8
3 26 24 10
4 37 35 12
5 50 48 14
6 65 63 16
7 82 80 18
8 101 99 20
9 122 120 22
10 145 143 24
Of course, family #2 contains the doubles of Family #1
There are an infinite number of such families. It might be interesting to derive a general formula for them.

Answer 3

I think I understand what he wants.

The numbers 3, 4, and 5 are a natural number solution to

a^2 + b^2 = c^2

When a = 3, b = 4, c = 5

So he is saying that if you were to give him the number 3 for the variable a, he could find two natural numbers for b and c that work in a^2 + b^2 = c^2.

What he wants to know is if a natural number c is given, can one find natural number solutions for a and b such that a^2 +b^2 = c^2

The answer is not always. In fact, natural number solutions to the pythagorean theorem are not easy to find.

Of course, one can always use ones that are known, like 3,4,5 and multiply each by a constant to get another triple (6, 8, 10 for example)

But say I gave you the number 12 for c and asked you to find two natural numbers for a and b such that a^2 + b^2 = c^2. In this case, I think you would find it impossible.

Answer 4

What do you mean by "natural solution" or "a" (natural)? This is strange terminology to my ears, and I've been studying math for 60 years. The PROBLEM is that we CAN'T follow the problem with wierd terminology left unexplained.

The equation you cite is a simple 3 variable one and you MUST have any 2 of them to determine the 3rd.

Answer 5

Hello Rashej

Yes we can, I think that you didnt read the answers on your question http://answers.yahoo.com/question/index;_ylt=Av5ZK1WdIlXXavpc2p9.nCXsy6IX?qid=20061104211451AALas6v , some of them are right about the pythagorean triplets with this you can solve given c, the a & b. The medthofd of ascending descent I do not know

So the answer is Yes we can with p triplets.

Answer 6

Common sense tells me you need to have three equations to solve for three variables.

Answer 7

you only have one equation, so you can only use it to solve for one variable. therefore you would need to know a and b to find c, a and c to find b, or c and b to find a. you can't find a and b by just knowing c.

Answer 8

Please Use the Pythogoras theorm or the Ascending decend to find the answer for U R question

Answer 9

naturally!

Answer 10

the question is all rubish. if the three (a,b,c) are unknown then u cannot solve it by only one equation.