True or False?
a) There exist a , b ∈ Z such that 2 0 1 9 = a 2 + b 2 .
a) There exist a , b ∈ Z such that 2 0 1 9 2 0 1 9 = a 2 + b 2 .
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Or, if there exists such a , b , knowing 2 0 1 9 = 3 × k , then a 2 + b 2 ≡ 0 m o d 3 should be true. However, according to Fermat little theorem, a 2 + b 2 ≡ 2 m o d 3 , given a , b have no factor of 3 . a , b have no factor of 3 , cause if they have then 2 0 1 9 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 :) ).
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2 0 1 9 ≡ 3 ( m o d 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.
2 0 1 9 2 0 1 9 ≡ 3 2 0 1 9 ( m o d 4 ) . The remainder of 3 n m o d 4 oscillates between 3 and 1 for odd and even values of n respectively. Since 2019 is odd, 3 2 0 1 9 ≡ 3 ( m o d 4 ) , so again there are no solutions.
Therefore, both statements are False .
Proof
Every number is either even or odd and can represented by 2 k or 2 k + 1 for some k . Squaring both of these cases gives
( 2 k ) 2 = 4 k 2 ≡ 0 ( m o d 4 )
( 2 k + 1 ) 2 = 4 k 2 + 4 k + 1 = 4 ( k 2 + k ) + 1 ≡ 1 ( m o d 4 )