Find all integers 'a' and 'b' that satisfy the equation: 4a2 + 4ab – 3b2 = 2029
2029 is prime.
One way to see this is to just try all of the primes up to `sqrt(2029)=45.04`
2029/2, 2029/3, ... 2029/43
None of these divides 2029, so 2029 is prime
Another way to see this is to consult a list of the first thousand primes, and check to see that 2029 is on the list.
2029 is prime, so its only divisors are 1 and 2029 (and -1, -2029)
So, we want to find a and b such that: `(2a+3b)(2a-b)=2029`
We have only 4 possibilities:
These are 4 systems of equations. We'll solve each of them, and see if the results give us integers for `a` and `b`
So this situation gives us integer values for a and b
Similarly, the other situations have solutions:
`2a+3b=2029`, `2a-b=1` `=> a=254, b=507`
`2a+3b=-1`, `2a-b=-2029` `=> a=-761, b=507'
`2a+3b=-2029`, `2a-b=-1` `=> a=-254 ,b=-507'
So: all the integers (a,b) that satisfy the equation are: