A lottery draw's numbers (1981, 44th game week) were as follows: 16, 25, 41, 64, 85. What is the probability that all the numbers in a lottery draw can be expressed as the sum of two square numbers? (0 is also considered a square number.)
A 90 lottery numbers can be drawn in $\binom{90}{5}$ ways. We will call those lottery numbers (numbers between 1 and 90) suitable if they can be expressed as the sum of two squares. If the number of these is $p$, then $\binom{p}{5}$ ways can result in all five drawn numbers being suitable. Since in the classical probability model we assume that the probability of every possible quintuple of numbers being drawn is the same, the sought probability is the ratio of the number of suitable cases to the number of all possible cases, that is, $\binom{p}{5} /\binom{90}{5}$. Now we only need to determine $p$, that is, the number of suitable cases. The number $n$ is suitable if it falls between 1 and 90 and is either of the form $n=b^{2}$ or $n=a^{2}+b^{2}$, where $a, b$ are integers, and $1 \leqq a \leqq b \leqq 9$.

Numbers of the form $n=b^{2}$ yield 9 suitable numbers: $1,4,9,16,25,36,49,64,81$.

Numbers of the form $n=1+b^{2}, 1 \leqq b \leqq 9$ yield 9 suitable numbers: $2,5,10,17,26,37,50,65,82$.

Numbers of the form $n=4+b^{2}, 2 \leqq b \leqq 9$ yield 8 suitable numbers: $8,13,20,29,40,53,68,85$.

Numbers of the form $n=9+b^{2}, 3 \leqq b \leqq 9$ yield 7 suitable numbers: $18,25,34,45,58,73,90$.

Numbers of the form $n=16+b^{2}, 4 \leqq b \leqq 9$ yield 5 suitable numbers: $32,41,52,65,80$.

Numbers of the form $n=25+b^{2}, 5 \leqq b \leqq 9$ yield 4 suitable numbers: $50,61,74,89$.

Numbers of the form $n=36+b^{2}, 6 \leqq b \leqq 9$ yield 2 suitable numbers: 72,85.

Numbers of the form $n=a^{2}+b^{2}, a, b \geqq 7$ do not yield any suitable numbers. Thus, there are 44 numbers, but four of them (25, 50, 65, 85) appear twice. Consequently, there are 40 distinct suitable numbers, that is, $p=40$. The sought probability is $\binom{40}{5} /\binom{90}{5} \approx 0.015$.

Remarks. 1. Out of 1295 draws between March 7, 1957, and the end of 1981, all five winning numbers were suitable 18 times. Based on the above result, in 1295 draws, this is expected to occur $1295 \cdot\binom{40}{5} /\binom{90}{5} \approx 19.3$ times.

2. A known theorem (see, e.g., Sárközy A.: Number Theory, Műszaki Könyvkiadó, 1976, p. 273) states that a natural number can be expressed as the sum of two squares if and only if the prime factors of the form $4k-1$ appear with even exponents in its prime factorization.