Estonian Math Competitions

Let the lengths of legs be $a$ and $b$. By the Pythagorean theorem, $$a^2+b^2=88\dots 822\dots 2.$$ If $k=1$ then one can choose $a=9$ and $b=1$ as $9^2+1^2=82$. If $k\ge 2$ then $88\dots 822\dots 2\equiv 6(\mathrm{mod}8)$ since $822\equiv 6(\mathrm{mod}8)$ and $222\equiv 6(\mathrm{mod}8)$. 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, $a^2+b^2=88\dots 822\dots 2$ cannot hold for $k\ge 2$.

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Alternative solution.

Let the lengths of legs be $a$ and $b$. By the Pythagorean theorem, $$a^2+b^2=88\dots 822\dots 2.$$ 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 $88\dots 822\dots 2=2\cdot 44\dots 411\dots 1$, 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 $44\dots 411\dots 1$, this number itself would be congruent to $1$ modulo $4$, but actually $44\dots 411\dots 1\equiv 3(\mathrm{mod}4)$. This shows that right triangles with the required property cannot exist in the case $k\ge 2$.