Circle on lattice
x 2 + y 2 = 2 0 1 5
How many integer solutions are there to the Diophantine equation above?
The answer is 0.
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4 solutions
Moderator note:
Thanks for explaining why the quadratic residues of 4 is 0 and 1 only.
Bonus question : Is there any integer solution(s) to x 2 + y 2 + z 2 = 2 0 1 5 ?
In response to Challenge Master :
There aren't any integer solutions to x 2 + y 2 + z 2 = 2 0 1 5 .
We can prove that quadratic residues of 8 is 0, 1 or 4. ( Hint : Prove that the perfect square of any integers can be written as 1 6 k 2 or 1 6 k 2 + 8 k + 1 or 1 6 k 2 + 1 6 k + 4 or 1 6 k 2 + 2 4 k + 9 ( k is an integer)).
Thus, the remainder of x 2 + y 2 + z 2 when divided by 8 can be only 0 , 1 , 2 , 3 , 4 , 5 , 6 , while 2 0 1 5 ≡ 7 ( m o d 8 ) .
Hey pal NYC soln but I am unfamiliar with quad residues can u explain me how you arrived at the answer 0 ( last step i dont get )if u can't fine atleast suggest me a good number theory book to start off with . Thnxs in adv
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Sorry but I don't really understand what do you mean by saying "the last step" is. Can you explain more specifically?
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It's been ages since I did math mostly these past three years I have been doing engg maths so I dunno how u used mod theory to do this prob. Okay fine now instead of 2015 rhs is 2016 then mod 4 will result in zero right .so now how many integral solutions are there ?
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@Rakesh Ramachandiran – And do me a favour dude suggest a real good simple to understand Number theory book with inspiring and motivating questions !
According to me the last remainder should be 1 because there are 4 possiblities that are (even,even);(even,odd);(odd,even);(odd,odd) . so the remainder of (even,odd) and (odd,even) should be 1.
The equation can be re-written as x 2 + y 2 − 2 0 1 5 = 0 . This is similar to x 2 + y 2 − 2 x y = ( x − y ) 2
For an integer solution the term 2 x y has to be an even number. Since it is not, it follows that there are no integer solutions.
But that would imply x = y , which is not necessarily true. For example, I could consider x 2 + y 2 = 1 , which would have 4 solutions in total; but x 2 + y 2 = 3 has no solutions.
Its just one step problem just apply Fermat's theorem on sums of two squares Therefore you just have to check whether 2015 is 1 mod 4 or not.And we found its not therefore no integral solution
A square number divided by 4 has a surplus of 0 or 1, so the sum of two square number divided by 4 has the surplus of 0 or 1 or 2. But 2015 divided by 4 has the surplus of 3. In conclusion that there is no interger number satisfied with the given equation.
Firstly, every integer must be either an odd or an even one. Hence, every integer n can be written as 2 k or 2 k + 1 ( k is an integer).
Thus, n 2 = 4 k 2 or n 2 = 4 k 2 + 4 k + 1 , which implies that n 2 ≡ 0 ( m o d 4 ) or n 2 ≡ 1 ( m o d 4 ) . The same method is applied to x 2 and y 2 to prove that both x 2 and y 2 are divisible by 4 or leave a remainder of 1 when divided by 4 .
So, x 2 + y 2 ≡ 0 ( m o d 4 ) or x 2 + y 2 ≡ 1 ( m o d 4 ) or x 2 + y 2 ≡ 2 ( m o d 4 ) or x 2 + y 2 ≡ 0 ( m o d 4 ) .
Meanwhile, 2 0 1 5 ≡ 3 ( m o d 4 ) .
This implies that there are 0 integer solutions to the equation.