Problem 1834

Let the lengths of legs be $a$ and $b$. By the Pythagorean theorem,
$$a^2 + b^2 = 88\ldots822\ldots2.$$
If $k=1$ then one can choose $a=9$ and $b=1$ as $9^2 + 1^2 = 82$. If $k \ge 2$ then $88\ldots822\ldots2 \equiv 6 \pmod{8}$ since $822 \equiv 6 \pmod{8}$ and $222 \equiv 6 \pmod{8}$. On the other hand, the residues of $a^2$ and $b^2$ modulo $8$ can be $0$, $1$, or $4$. However, the sum of such two numbers can have residue $0$, $1$, $2$, $4$, or $5$ modulo $8$. Consequently, $a^2 + b^2 = 88\ldots822\ldots2$ cannot hold for $k \ge 2$.

Let the lengths of legs be $a$ and $b$. By the Pythagorean theorem,
$$a^2 + b^2 = 88\ldots822\ldots2.$$
If $k=1$ then one can choose $a=9$ and $b=1$ as $9^2 + 1^2 = 82$. Assume in the rest that $k \ge 2$. It is known that a positive integer is expressible as the sum of two squares of integers if and only if its canonical representation contains primes congruent to $3$ modulo $4$ only with even exponents. Note that $88\ldots822\ldots2 = 2 \cdot 44\ldots411\ldots1$, where only the odd second factor can be divisible by primes congruent to $3$ modulo $4$. If all such primes would occur with even exponents in the canonical representation of $44\ldots411\ldots1$, this number itself would be congruent to $1$ modulo $4$, but actually $44\ldots411\ldots1 \equiv 3 \pmod{4}$. This shows that right triangles with the required property cannot exist in the case $k \ge 2$.