2019, ouhh year

$2019 \equiv 3 \pmod 4$ , but square numbers are congruent to 0 or 1 (proof below), so the sum of two square numbers can't be congruent to 3 modulo 4.

$2019^{2019} \equiv 3^{2019} \pmod 4$ . The remainder of $3^n \mod 4$ oscillates between 3 and 1 for odd and even values of $n$ respectively. Since 2019 is odd, $3^{2019} \equiv 3 \pmod 4$ , so again there are no solutions.

Therefore, both statements are False .

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Proof

Every number is either even or odd and can represented by $2k$ or $2k+1$ for some $k$ . Squaring both of these cases gives

$(2k)^2 = 4k^2 \equiv 0 \pmod 4$

$(2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2+k)+1 \equiv 1 \pmod 4$

Or, if there exists such $a,b$ , knowing $2019=3\times k$ , then $a^2+b^2 \equiv 0 \mod \ 3$ should be true. However, according to Fermat little theorem, $a^2+b^2 \equiv 2 \ \mod \ 3$ , given $a,b$ have no factor of $3$ . $a,b$ have no factor of $3$ , cause if they have then $2019$ should have an even power of $3$ , which is not true. All the trouble to conclude there exist no such $a,b$ .

Same can be done for part "b" (which is mistakenly "a" again :) ).