## Task 17/82

Find all two-digit natural numbers $n$ that are equal to the square of their cross sum.
Let $n=10 a_{1}+a_{2}$ with $0 \leq a_{1} ; a_{2} \leq 9, a_{1} ; a_{2} \in N$ be such a number and $q=a_{1}+a_{2}$ its digit sum. Then, because $a_{2}=q-a_{1}$,

$$
q^{2}=10 a_{1}+a_{2}=9 a_{1}+q \quad \text { so } \quad q^{2}-q=q(q-1)=9 a_{1}
$$

The product of two consecutive natural numbers is divisible by 9 if and only if one of the factors is divisible by 9. Since further $0 \leq a_{1} \leq 9$ holds, 

$$
q=9, q-1=8=a_{1}, a_{2}=q-a_{1}=1 \quad \text { or } \quad q=1, q-1=0=a_{1}, a_{2}=q-a_{1}=1
$$

$q=0, a_{1}=0, a_{2}=0$ is ruled out as trivial. The check confirms that $n_{1}=81$ and $n_{2}=01$ are indeed solutions to the problem (although $n_{2}=01$ will likely be excluded as not "genuinely" two-digit).