Another diophantine equation

Subtracting the second of the given equations from the first yields the equation

$x^{2} - x - (y^{2} - y) = 24 \Longrightarrow (x - \frac{1}{2})^{2} - (y - \frac{1}{2})^{2} = 24 \Longrightarrow$

$(2x - 1)^{2} - (2y - 1)^{2} = 96 \Longrightarrow (2x - 1 - (2y - 1))(2x - 1 + (2y - 1)) = 96 \Longrightarrow$

$(2(x - y))(2(x + y - 1)) = 96 \Longrightarrow (x - y)(x + y - 1) = 24$ .

Now $24 = 2^{3} \times 3$ has $(3 + 1)(1 + 1) = 8$ positive divisors, and thus $16$ integer divisors. We can then assign $(x - y)$ and $(x + y - 1)$ values in $16$ different ways. For example, we could have

$x - y = -24, x + y - 1 = -1 \Longrightarrow (x - y) + (x + y - 1) = -24 + (-1) \Longrightarrow$

$2x - 1 = -25 \Longrightarrow x = -12, y = 12, z = x^{2} + y - 42 = 114$ .

Similarly $x - y = -1, x + y - 1 = -24 \Longrightarrow x = -12, y = -11, z = 91$ .

However, for $x - y = -12, x + y - 1 = -2 \Longrightarrow 2x - 1 = -14$ , we do not obtain an integer value for $x$ . Only in those cases where the factor pairs add to an odd number will we get integer values for $x,y$ . There are 8 such factor pairings, including the pair $-24 \times -1$ mentioned above. The other 7 are

$-1 \times -24, -3 \times -8, -8 \times -3, 3 \times 8, 8 \times 3, 1 \times 24$ and $24 \times 1$ .

Each of the 8 factor pairings yields a distinct integer triple $(x,y,z)$ , so the given system of equations has a total of $\boxed{8}$ solutions.

Solve for $z$ in two different ways to simplify the equation into $x^2 - x - (y^2 - y) = 24$ . Divide by two on both sides: $\frac{x^2 - x}{2} - \frac{y^2 - y}{2} = 12$ . What we have is that the difference of two triangular numbers is equal to 12. But we know that the difference of two triangular numbers is always the sum of consecutive positive integers so we must ask, in how many ways can we write 12 as a sum of consecutive integers?. It is just an easy computation, 12 = 3 + 4 + 5. No other solutions exist. To which triangular numbers do these correspond? Well, $12 = 12$ just corresponds to the difference $\sum_{k=1}^{12} k - \sum_{k=1}^{11} k$ . So $T_x = 78, T_y = 66$ . It is important to note that when solving for the values of $x,y$ we will be solving quadratic equations with two distinct solutions, so we know this will generate $2 \times 2 = 4$ possible solutions.

To conclude, $12 = 3 + 4 + 5$ corresponds to the difference $\sum_{k=1}^{5} k - \sum_{k=1}^2 k$ . That is, $T_x = 15, T_y = 3$ . As before, this will generate 4 more solutions for a grand total of 8.

When we subtract equation 2 from 1 we get this after simplifying the result (x-y)(x+y-1)=24

If we factorize 24 into product of 2 numbers we get 8 different possible pairs of numbers they are:-

(1,24) (24,1) (2,12) (12,2) (3,8) (8,3) (4,6) (6,4) in these pairs we can allot these values to any one of the two terms in x,y [x-y and x+y-1] and get 8 values for x and y. As x,y are integers from the solutions we can say z is also integer by closure property and thats it its done.

Ans: 8