Okay, here is the proof that no integers a and b will satisfy the equation a^2 = 3b^2 + 2015:
1) We can rewrite the LHS as a^2 = a * a, where a is an integer. The RHS (3b^2 + 2015) is a multiple of 3 plus 2015. So a * a must also be a multiple of 3 plus 2015 for any integer solutions.
2) The only perfect squares that fit this form are 0, 3, 12, 21, 30, 39, 48, 57, 66, 72, 79, 84, 87, 90, 99, ... (multiples of 3 plus 2015). None of these are equal to 3b^2 for any integer b.
3) Let's suppose a = 3k for some integer k. Then 9k^2 = 3b^2 + 2015. But 3 does not divide 9k^2 except when k = 0, and 0^2 does not equal 3b^2 + 2015 for any b. Contradiction.
4) Let's suppose a = 3k+1 for some integer k. Then (3k+1)^2 = 3b^2 + 2015. But (3k+1)^2 is always 1 more than a multiple of 3, while 3b^2 + 2015 is a multiple of 3 plus 2015. Contradiction.
5) Let's suppose a = 3k+2 for some integer k. Then (3k+2)^2 = 3b^2 + 2015. But (3k+2)^2 is always 4 more than a multiple of 3, while 3b^2 + 2015 is a multiple of 3 plus 2015. Contradiction.
In all cases, we reach a contradiction. Therefore, there are no integer solutions for a and b that satisfy the original equation a^2 = 3b^2 + 2015.
Let me know if any part of this proof is unclear! I can provide more details or examples if needed.