Homogeneous Quadratic

The equation simplifies to $x^2 + (x - y)^2 = 225$ , so $y = x \pm \sqrt{225 - x^2}$ . Since $y$ is real, thus $-15 \leq x \leq 15$ . Now we try all possible integer values of $x$ within this range to see which ones give an integer value of $y$ . the integer values which give positive results are $15, -15, 12, -12, 9, -9$ and $0$ . Of these 15 and -15 give one possible value of y each, while the rest give two possible values of y each. thus we have in total 12 pairs of $(x, y)$ satisfying the equation.

Rewrite the formula as x^2 + (x-y)^2 = 15^2 Now the problem becomes how to express 15^2 as the sum of two integer squares. Start with x = 15 and x-y = 0. Try x = 14, 13, etc. down to 0. We find three more pair of squares this way, namely 12^2+9^2, 9^2+12^2 and 0^2 + 15^2. For the two pairs with zeroes, we get twice their number by negating the nonzero number, that's four pairs. For the non-zero pairs, we can assign negative signs to each member of the pair independently, so we get four times their number = eight pairs. Total is twelve.

After modification, the above relation becomes x^2 + (y-x)^2=225. now , let x=k (say) and put in above equation and find the value of y as: y=k \pm sqrt{225-k^2}. i.e. 1. y=k+ sqrt{225-k^2} and 2. y=k- sqrt{225-k^2}

Now take the equation 1 and proceed as below: since, y to be real -15 \leq k \leq 15 (k belongs to [-15, 15] ) then put the integral value of k from -15 to 15 . Putting k=x =-15, 15, -12, 12, -9, 9. we get the values of y as an integer. the solution for equation(1) is (15, 15), (-15, -15), (12, 21), (-12, -3), (-9, 12), (9, 21)..

similarly , the solution of equation(2).

Finally, we get the pairs of integers as SOLUTION.

Thanks.

LHS can be written as (x-y)^2 + x^2 RHS is 225

The possibilities for 225 are (0,15) and (9,12)

Next we substitute values for x from the above options and solve for values of y

For (0,15)

x= 0 we get y = 15 and -15 x = 15 we get y = 15 x = -15 we get y = -15 Therefore 4 possibilities

x = 9 we get -3 and 21 Similarly for x = -9 , 12,-12 we get 2 solutions each Therefore 8 possibilities

Total of 12 possibilities

x^2+{x-y}^2=15^2 x=12x-y=9and negatives X=9 x-y=12 x=0 x-y =15 and negatives 3*4=12

We have $x^2 + (x-y)^2 = 15^2$ . A quick check reveals that $15^2 = 0^2 +(\pm 15)^2 = (\pm 9)^2 + (\pm 12)^2$ are the only ways to write $15^2$ as the sum or 2 integer squares. So $(x, x-y) = (0, \pm 15), (\pm 15, 0 ), (\pm 9, \pm 12), (\pm 12, \pm 9)$ and there are $2 + 2 + 4 + 4= 12$ pairs of $(x,y)$ .

Notice that the equation implies $x,y \equiv 0 \mod 3$ . This is because if we assume neither is divisible by 3, this implies $-2xy \equiv 0 \mod 3$ which is a contradiction. If we then assume that one of them is divisible by 3 then we will have $2x^2 \equiv 0 \mod 3$ or $y^2 \equiv 0 \mod 3$ which implies the other is also divisible by 3. So do the substitution $x = 3k, y=3q$ . Because the equation is homogeneous we can divide both sides by $9$ to get a new equation:

$2k^2 - 2kq + q^2 = 25$ . This new equation is easier to solve and it has the property that $(k,q)$ is a solution of this equation if and only if $(3k,3q)$ is a solution of the original equation, which helps as we only care about counting solutions. Then we have the factorization:

$k^2 + (k-q)^2 = 5^2$ . Let's now divide into cases:

Case 1: $k=0$ . In this case we would need to have $(k-q)^2 = 25 \implies (-q)^2 = 5 \implies q = \pm 5$ . These are two solutions

Case 2: $k-q = 0$ . Then we would need $k^2 = 25 \implies k = \pm 5$ . Two more solutions.

Case 3: Neither $k$ or $k-q$ are equal to 0. We would now be operating in usual Pythagorean triples. We know that $3^2 + 4^2 = 5^2$ is the only solution (ignoring changes in sign) so in this step one would choose $k = 3,-3,4,-4$ and then solve the resulting quadratic equation for $q$ . Example: Let $k = 3$ . Then necessarily $(k-q)^2 = 16$ so $(3-q) = \pm 4 \implies q =3 \pm 4 = 7,-1$ .

Repeat this process until you have exhausted the four options for $k$ and you'll get $8$ more solutions. Adding up to a total of 12.