**
True or False?
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a) There exist $a, b \in \mathbb{Z}$ such that $2019 = a^2 + b^2$ .

a) There exist $a, b \in \mathbb{Z}$ such that $2019^{2019} = a^2 + b^2$ .

a)False b)True
a)True, b) True
a)False b)False
a)True, b) False

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$2019 \equiv 3 \pmod 4$ , 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.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

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

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