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The number of integer solutions for the equation $x^2+y^2=2011$ is

  1. $0$
  2. $1$
  3. $2$
  4. $3$
in Numerical Ability by Veteran (425k points)
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Answer: $A$

Remainder when $2011$ is divided by $4  = 3$

Now, we know that the square of any even number is divisible by $4$.

$$\implies (2k)^2 = 4 \times k^2,\;where\;k\in Even\;number$$

Also, square of any odd number gives the remainder $1$ when divided by $4$:

$$\implies (2k+1)^2 = 4k^2 + 4k + 1= 4 \times (k^2 + k ) + 1$$

Thus the square of any integer when divided by $4$ either gives $0$ or $1$ as the remainder.

Now, when we add $x^2 + y^2 $ together.

Possible remainders $ = 0+0 = 0, \;0+1= 1, \;1 + 0 = 1, \; 1 + 1 = 2 = \bf\{0, 1, 2\}$

So, we will never get $3$ as the remainder.

Hence, the number of integer solutions for the above equation $=0$

$\therefore A$ is the right option.
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