Q13/127) find integers in positive solutions to xyz + 20xy + yz + 5zx + 100x + 20y + 5z = 1913

Because we have xyz, xy, yz, xz, x, y, z terms in the left we should add some d and factor if possible

(x+a)(y+b)(z+c) – d

= xyz + ayz + bzx + cxy + bcx + abz + acy + abc – d

Comparing with given expression we get a = 1,  b = 5, c = 20

So we get (x+1)(y+5)(z+20) = xyz+ yz +20xy + 5xyz + 100 x + 5z + 20y + 100

Comparing with coefficient we get d = 100

So (x+1)(y+5)(z+20) = 1913 + 100 = 2013 = 3 * 11 * 61

So x = 2, y = 6, z= 41