Question:- How many numbers are there between 1 to 1000 which can be expressed by the difference of two square numbers?
Pls help to solve it.i have used brute force but getting wrong answer:(
Here is my Code in C++
The output is 327,as there is repetition i have used set to store the unique numbers only.
#include<iostream>
using namespace std;
int main()
{
int a[32];
set<int>s;
for(int i = 1;i <= 31;i++){
a=i*i;
}
for(int i = 1;i <= 30;i++){
for(int j = i+1;j <= 31;j++){
s.insert(abs(a-a[j]));
cout<<abs(a-a[j])<<endl;
}
}
cout << s.size() <<endl;
}
}
Number Theory
- Anindya Biswas
- Posts:264
- Joined:Fri Oct 02, 2020 8:51 pm
- Location:Magura, Bangladesh
- Contact:
Re: Number Theory
If you want to post about programming or other computer science related topic, you should post here.
For number theory, post here.
Every odd number is difference of two squares, $2n+1=(n+1)^2-n^2$
Every even number that's divisible by $4$ is difference of two squares, $4n=(n+1)^2-(n-1)^2$.
Every number of the form $4n+2$ can't be a difference of two squares, for the sake of contradiction, let's assume,
$4n+2=a^2-b^2=(a+b)(a-b)$
Now, $a+b\equiv a-b\pmod{2}$
So, $a+b,a-b$ both must be even. Therefore, $4\mid (a+b)(a-b)$.
But $4\nmid 4n+2$, which is a contradiction.
So, use this code instead,
For number theory, post here.
Every odd number is difference of two squares, $2n+1=(n+1)^2-n^2$
Every even number that's divisible by $4$ is difference of two squares, $4n=(n+1)^2-(n-1)^2$.
Every number of the form $4n+2$ can't be a difference of two squares, for the sake of contradiction, let's assume,
$4n+2=a^2-b^2=(a+b)(a-b)$
Now, $a+b\equiv a-b\pmod{2}$
So, $a+b,a-b$ both must be even. Therefore, $4\mid (a+b)(a-b)$.
But $4\nmid 4n+2$, which is a contradiction.
So, use this code instead,
Code: Select all
include <iostream>
include <cmath>
using namespace std;
int main()
{
int n=1000;
int a=ceil(n/2);
int b=floor(n/4);
return a+b;
}
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann