x2+y2= 2013
How many solutions (x,y) exist such that x and y are positive integers?
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2 Answers
23052013 = 3*271
271 and 3 are of the form 4k+3 while the required condition for a number n to be a sum of 2 squares is that:
n = 2ma2b
where m≥0,a is odd, and b and every prime divisor of b is of the form 4k+3.
Since 2013 does not satisfy the conditions no solution exist.
Swarna Kamal Dhyawala it should be 2013= 671*3!!!!!!!!!!Upvote·0· Reply ·2013-02-08 10:11:16- Swarna Kamal Dhyawala 2013 =671*3
Soumyabrata Mondal if the number is nt power of any no then wot??- Soumyabrata Mondal sorry I mean if the no is prime then wot??
- Shaswata Roy Sorry, b should have prime divisors of the form 4k+1
2305In the previous answer I mentioned that b should have prime divisors of the form 4k+3 when it should actually be 4k+1.
and,2013 = 3*11*61.
@Swarna Kamal Dhyawala:Thanks for rectifying the mistake.
2013 = 20*12*2013
Here m=0, a =1, and b =2013 = 3*11*61
Since all the prime divisors of b are not of the form 4k+1,it cannot be written as a sum of 2 squares.
@Soumyabrata Mondal: If the number is prime it would mean that m =0,a =1 and b is a prime.
Now if b is not of the form 4k+1 it cannot be written as a sum of 2 squares.Otherwise Solutions to x and y exist.