please elaborate the statement
Did not understand statement
@deepgarg46 By statement you meant that you are not able to understand question?
If yes :
Then in this question you will be given a no. n and you have to find the Pythagoras triplets for that no. for example : if n = 3 then you need to find two no. that will satisfy the condition 3^2 +b^2 = c^2; you need to find b and c and remember 3 < b < c should be considered.To find this you need to use a simple trick formula give below :
- If a is even, then let’s try to equate it with the second cathetus, 2mn, i.e., let’s try to find an (m,n) pair such that 2mn = a. In this case, is even, so we can divide by 2. We have mn = a/2. Thus, all we have to do is to find a factorization of a/2 into mn such that m>n. You can try various factoring methods — even naïve O(√a) time method will pass - but in this case , we don’t really even need to do that, since the trivial (m,n) = (a/2,1) works!. This gives us the solution triple ( (a^2)/4 - 1, a, (a^2)/4 + 1 ). Note that (a^2)/4 is an integer since a is even.
- If a is odd, then we can’t equate it with the second cathetus, so let’s try instead m2-n2, i.e. we want to find an (m,n) such that m2-n2 = a. Node that m2-n2 = (m+n)(m-n), so all boils down again to factoring! Let’s try our trivial factorization again: (m+n, m-n) = (a,1) What we get is a system of two linear equations: m+n=a and m-n=1 which has solution m=(a+1)/2 and n=(a-1)/2. Substituting back, this gives us the triple (a, (a^2 - 1)/2, (a^2 + 1)/2). Note that (a^2 + 1)/2 is an integer as a is odd.
Refer the code below :
I hope I’ve cleared your doubt. I ask you to please rate your experience here
Your feedback is very important. It helps us improve our platform and hence provide you
the learning experience you deserve.
On the off chance, you still have some questions or not find the answers satisfactory, you may reopen
the doubt.