If a and b be any two positive integers which on division by 6 earn remainder…

When two numbers are combined by addition, their remainders modulo m combine the same way; and if a number leaves remainder r on division by m, its square leaves the same remainder as r2 on division by m. So the remainder of a2 + b2 on division by 6 can be found by squaring each given remainder, reducing modulo 6, and adding the results.

1. a leaves remainder 4 on division by 6, so a2 leaves the same remainder as 42 = 16 on division by 6; since 16 = 2 × 6 + 4, a2 ≡ 4 (mod 6).

2. b leaves remainder 5 on division by 6, so b2 leaves the same remainder as 52 = 25 on division by 6; since 25 = 4 × 6 + 1, b2 ≡ 1 (mod 6).

3. Adding the two partial remainders gives a2 + b2 ≡ 4 + 1 = 5 (mod 6).

4. Since 5 is already less than 6, no further reduction is needed, so the remainder of a2 + b2 on division by 6 is 5.

As a check, take concrete values satisfying the given conditions: a = 10 (since 10 = 6 × 1 + 4) and b = 11 (since 11 = 6 × 1 + 5). Then a2 + b2 = 100 + 121 = 221, and 221 = 6 × 36 + 5, which again gives a remainder of 5.

Also note that a remainder from division by 6 always lies in {0, 1, 2, 3, 4, 5}, so 6 itself could never be a valid remainder for this division.

So the remainder of a2 + b2 on division by 6 is 5.