Let a and b be integers (not necessarily positive). Prove that (a^3 + 5b^3) not equal to 2016.
To prove that (a^3 + 5b^3) is not equal to 2016 for any integers a and b, we can begin by assuming the opposite: suppose there exist integers a and b such that (a^3 + 5b^3) = 2016.
We can rearrange the equation to obtain: a^3 = 2016 - 5b^3.
Now, let's consider the cube of any integer. The possible remainders when dividing a cube by 5 are 0, 1, -1, 2, or -2. This can be checked by testing cubes of integers:
0^3 = 0 (remainder 0 when divided by 5)
1^3 = 1 (remainder 1 when divided by 5)
2^3 = 8 (remainder 3 when divided by 5)
3^3 = 27 (remainder 2 when divided by 5)
4^3 = 64 (remainder 4 when divided by 5)
...
From this, we can observe that the cubes of integers can only have remainders 0, 1, -1, 2, or -2 when divided by 5. Knowing this, we can determine the possible remainders for a^3 - 5b^3 in the equation a^3 = 2016 - 5b^3:
2016 - 5b^3 ≡ 0, 1, -1, 2, or -2 (mod 5)
Since 2016 is divisible by 5 (the remainder is 0 when divided by 5), we conclude that the left side of the equation (a^3 - 5b^3) must also be divisible by 5. Therefore, the right side of the equation (2016) must leave a remainder of 0 when divided by 5.
However, if we examine the remainders of the first few positive integers modulo 5, we see that none of them are equal to 0 when squared:
0 ≡ 0 (mod 5)
1 ≡ 1 (mod 5)
2 ≡ 2 (mod 5)
3 ≡ 3 (mod 5)
4 ≡ 4 (mod 5)
Since no integer squared leaves a remainder of 0 when divided by 5, we have a contradiction since a^3 cannot be equal to 0 when divided by 5. Therefore, our initial assumption that (a^3 + 5b^3) = 2016 is false.
Therefore, we have proven that (a^3 + 5b^3) is not equal to 2016 for any integers a and b.