How to prove if a,b,c,d are integers

if a,b,c,d are integers with p =√(a²+b²),q=√[(a-c)²+(b-d)²] and r = √(c²+d²)
prove that √[(p+q+r)*(p+q-r)*(p-q+r)*((-p)+q+r)] is even integer?

The numbers p, q, and r are lengths of the sides of a triangle with vertices (0, 0), (a, b), (c, d), and, by Heron's formula, the number presented is 4 times the area of the triangle. So, this is equivalent to proving that this triangle has half of an integer area, for any integers a, b, c, d.

Well, if we take the segment between (0, 0) and (a, b) to be the base, then the perpendicular height will be the length of the orthogonal component of (c, d) with respect to (a, b). The parallel projection is:

((a, b).(c, d) / ||(a, b)||²)(a, b)

and so the orthogonal component is:

(c, d) - ((a, b).(c, d) / ||(a, b)||²)(a, b)
= (c, d) - (ac + bd) / (a² + b²))(a, b)
= (c - a(ac + bd) / (a² + b²), d - b(ac + bd) / (a² + b²))
= (c(a² + b²) - a(ac + bd), d(a² + b²) - b(ac + bd)) / (a² + b²)
= (b²c - abd, a²d - abc) / (a² + b²)

The norm of this orthogonal component is the height of the triangle, which is given by:

√((b²c - abd)² + (a²d - abc)²) / (a² + b²)
= √(b⁴c² - 2ab³cd + a²b²d² + a⁴d² - 2a³bcd + a²b²c²) / (a² + b²)
= √(b⁴c² + a²b²c² + a⁴d² + a²b²d² - 2abcd(a² + b²)) / (a² + b²)
= √(b²c²(a² + b²) + a²d²(a² + b²) - 2abcd(a² + b²)) / (a² + b²)
= √(b²c² + a²d² - 2abcd) / √(a² + b²)
= √(bc - ad)² / √(a² + b²)
= |bc - ad| / √(a² + b²)

So, the area, given by half times the base times the height, is:

(1/2) * |bc - ad| / √(a² + b²) * ||(a, b)||
= (1/2) * |bc - ad| / √(a² + b²) * √(a² + b²)
= (1/2) * |bc - ad|

So, the area of the triangle is given by |bc - ad| / 2, which is half of an integer, which is what we wanted to prove. Hence the number you gave is an even integer.

It takes some effort, but you can show that the expression under the radical is equal to

4(ad - bc)²

So that the final result is the even integer 2|ad - bc|.

Note that

(p+q + r)(p + q - r)(r + p - q)(r - (p - q)) =

((p+q)² - r²)(r² - (p - q)²) =

r²((p+q)² + (p - q)²) - r^4 - (p² - q²)² =

2p²r² + 2q²(r² + p²) - (q^4 + r^4 + p^4) =

4p²r² - (q² - (r² + p²))² =

4(a² + b²)(c² + d²) - (a² + b² + c² + d² - 2ac - 2bd - a² - b² - c² - d²)² =

4(a² + b²)(c² + d²) - 4(ac + bd)² =

4[a²c² + b²d² + a²d² + b²c² - a²c² - b²d² - 2abcd] =

4(ad - bc)².