An algebra problem by Kashish Vashist

$x^2 + y^2$ can only be divisible by 5 when

Case 1: $x^2$ and $y^2$ are both divisible by 5

Case 2: $x^2$ leaves a remainder of 1 and $y^2$ leaves a remainder of 4 when divided by 5

Case 3: $x^2$ leaves a remainder of 2 and $y^2$ leaves a remainder of 3 when divided by 5

Case 4: $x^2$ leaves a remainder of 3 and $y^2$ leaves a remainder of 2 when divided by 5

Case 5: $x^2$ leaves a remainder of 4 and $y^2$ leaves a remainder of 1 when divided by 5

For case 1: For any square to be divisible by 5, its root must also be divisible by 5.And the probability of a number being divisible by 5 is $\frac{1}{5}$

$\therefore$ The probability of $x^2$ and $y^2$ to be divisible by 5 is $\frac{1}{5^2} = \frac{1}{25}$

For case 2:For any number to leave a remainder of 1 when divided by 5, it must have 1 or 6 as its unit digit.And the probability of a square having 1 or 6 as its unit digit is $\frac{4}{10}$ , since there are 4 digits(1,4,6,9) which when squared, give the the units digit as 1 or 6.

And for any number to leave remainder 4 when divided by 5, it must have its unit digit 4 or 9. And the probability for such is $\frac{4}{10}$ , since there are 4 digits(2,3,7,8), which when squared, give 4 or 9 as the units digit.

So the probability of case 2 = $\frac{4}{10} \times \frac{4}{10} = \frac{4}{25}$

For case 3=For any number to leave a remainder of 2 when divided by 5 , it must have its unit digit as 2 or 7. But since no square can have those digits at the units place, the probability of a square having 2 or 7 as its unit digit is 0

For any number to leave a remainder of 3 when divided by 5 , it must have its unit digit as 3 or 8. But since no square can have those digits at the units place, the probability of a square having 3 or 8 as its unit digit is 0

So the probability of case 3 is 0

For case 4: Case 4 is just the inverse of case 3.

$\therefore$ The probability of case 4 is 0

For case 5: Case 5 is just the inverse of case 2

$\therefore$ The probability of case 5 is $\frac{4}{25}$

$\therefore$ The probability of $x^2 + y^2$ to be divisible by 5 is

$\frac{1}{25} + \frac{4}{25} + \frac{4}{25} = \boxed{\frac{9}{25}}$

CONSIDER FIRST 5 NATURAL NUMBERS SO COMBINATIONS ARE (1,1) (1,2) (1,3) (1,4) (1,5)........................... SO 25 POSSIBILITY OUT OF WHICH 9 OF THEM SATISFIES SO 9/25. SO INFINITE NUMBERS WILL ALSO IN SAME RATIO