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Two natural numbers $x$ and $y$ are selected at random. Find the probability that $x^2 + y^2$ is divisible by $10$.

I have no clue how to do this.

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    consider uniquely the last number of $x$ and $y$, that reduce the possibilities to $0<=x,y<=9 – 2017-01-10
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    For a new answer to the original question see there: http://math.stackexchange.com/questions/271437/probability-of-two-integers-square-sum-divisible-by-10. – 2017-01-10

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We only need to see the last digit of squares of both numbers which should add up to 10 for the number to be divisible by 10. Below are the possible last digits after squaring a number :

$$last\space digit(..0^2) = 0$$ $$last\space digit(..1^2) = 1$$ $$last\space digit(..2^2) = 4$$ $$last\space digit(..3^2) = 9$$ $$last\space digit(..4^2) = 6$$ $$last\space digit(..5^2) = 5$$ $$last\space digit(..6^2) = 6$$ $$last\space digit(..7^2) = 9$$ $$last\space digit(..8^2) = 4$$ $$last\space digit(..9^2) = 1$$ Now for sum of squares of 2 nos. to be divisible by $10$, the possible combinations will be $${(..0^2+..0^2)}$$ $${(..1^2+..3^2)}$$ $${(..1^2+..7^2)}$$ $${(..2^2+..4^2)}$$ $${(..2^2+..6^2)}$$ $${(..3^2+..9^2)}$$ $${(..4^2+..8^2)}$$ $${(..5^2+..5^2)}$$ $${(..6^2+..8^2)}$$ $${(..7^2+..9^2)}$$ $$No.\space of\space favorable\space cases = 18$$ since values of $x$ and $y$ can be swapped except when $x = y$.

Now $$Total\space Cases = 10\cdot 10 = 100$$ $$Required\space Probability = \frac {18}{100}$$

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You have to define what it means "at random" on natural numbers. It cannot be uniform probability (all numbers can be selected with equal probability), since it is not defined on the natural numbers.

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    Ocam's razor suggests to assume that both $x$ and $y$ are independently and uniformly distributed mod $10$. – 2017-01-10
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    @Christian Blatter Why not choose them independent with geometric distribution? Or choose them in the naive way, which is uniform on $\{1,\ldots n\}$ and let $n\rightarrow\infty$? – 2017-01-11