Find positive integer solutions

[anemone]

Find all positive integer solutions to:

$xyzt-2xyz-xyt+2xy+zt-2z-t=12$

 

[MarkFL]

My solution:

Spoiler

If we add 2 to both sides, and factor, we obtain:

\(\displaystyle (t-2)(z-1)(xy+1)=14\)

Looking at the permutations of the factorization $14=1\cdot1\cdot14$, we find:

i) \(\displaystyle t-2=1\implies t=3\)

\(\displaystyle z-1=1\implies z=2\)

\(\displaystyle xy+1=14\implies xy=13\implies (x,y)=(1,13),\,(x,y)=(13,1)\)

ii) \(\displaystyle t-2=1\implies t=3\)

\(\displaystyle z-1=14\implies z=15\)

\(\displaystyle xy+1=1\implies xy=0\) This case doesn't meet the requirements given.

The third permutation won't work for the same reason.

Looking at the permutations of the factorization $14=1\cdot2\cdot7$, (where the third factor isn't 1) we find:

i) \(\displaystyle t-2=1\implies t=3\)

\(\displaystyle z-1=2\implies z=3\)

\(\displaystyle xy+1=7\implies xy=6\implies (x,y)=(1,6),\,(6,1),\,(2,3),\,(3,2)\)

ii) \(\displaystyle t-2=1\implies t=3\)

\(\displaystyle z-1=7\implies z=8\)

\(\displaystyle xy+1=2\implies xy=1\implies (x,y)=(1,1)\)

iii) \(\displaystyle t-2=2\implies t=4\)

\(\displaystyle z-1=1\implies z=2\)

\(\displaystyle xy+1=7\implies xy=6\implies (x,y)=(1,6),\,(6,1),\,(2,3),\,(3,2)\)

iv) \(\displaystyle t-2=7\implies t=9\)

\(\displaystyle z-1=1\implies z=2\)

\(\displaystyle xy+1=2\implies xy=1\implies (x,y)=(1,1)\)

Thus, we find the following 12 solutions:

\(\displaystyle (t,x,y,z)=(3,1,13,2)\)

\(\displaystyle (t,x,y,z)=(3,13,1,2)\)

\(\displaystyle (t,x,y,z)=(3,1,6,3)\)

\(\displaystyle (t,x,y,z)=(3,6,1,3)\)

\(\displaystyle (t,x,y,z)=(3,2,3,3)\)

\(\displaystyle (t,x,y,z)=(3,3,2,3)\)

\(\displaystyle (t,x,y,z)=(3,1,1,8)\)

\(\displaystyle (t,x,y,z)=(4,1,6,2)\)

\(\displaystyle (t,x,y,z)=(4,6,1,2)\)

\(\displaystyle (t,x,y,z)=(4,2,3,2)\)

\(\displaystyle (t,x,y,z)=(4,3,2,2)\)

\(\displaystyle (t,x,y,z)=(9,1,1,2)\)

 

[anemone]

Very well done, MarkFL! Thanks for participating!