# Find all integers 'a' and 'b' that satisfy the equation: 4a2 + 4ab – 3b2 = 2029

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### 1 Answer

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)

Now, `4a^2+4ab-3b^2=(2a+3b)(2a-b)`

So, we want to find a and b such that: `(2a+3b)(2a-b)=2029`

We have only 4 possibilities:

`2a+3b=1`, `2a-b=2029`

`2a+3b=2029`, `2a-b=1`

`2a+3b=-1`, `2a-b=-2029`

`2a+3b=-2029`, `2a-b=-1`

These are 4 systems of equations. We'll solve each of them, and see if the results give us integers for `a` and `b`

`2a+3b=1`

`2a-b=2029`

`4b=-2028`

` b=-507`

`a=761`

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:

(761,-507)

(254,507)

(-761,507)

(-254,-507)