A Very Nice Polynomial System

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x+y=5, xz+yt=7, ...
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[devondevon]

Answer x=3 and y=2 and in this z=1 and t=2
or x=2 and y =3 and in this case, z=2 and t=1
or (3, 2) (2, 1) or (2, 3) (1, 2)
If assumed the values are positive integers, starting with t^3, t cannot be > 3 since 3^3 = 27. The same
can be said of z^3. Hence, for x+ y = 5, either x or y =2: hence x or y =3.
So assuming t =2, then t^3 = 8, and y=2; hence yt^3 =16; hence xz^3 = 3; since x =3
then z^3 = 1 hence z =1
Hence x=2 or 3, y= 2 or y , z = 1 or 2

[guyvankerckhoven]

My first guess (and guessing is quite easy for this problem) turned out to be correct. But it is great to see the way you solve it👍👍

[allanmarder]

I did this by guess and check. The first column sum is x*(1+ z + z^2 +z^3) The second column sum is y*(1 + t + t^2 + t^3)
The sum of the 1st 2 rows is 12 the sum of the second 2 rows is 30 so I tried to see if I could get the 2 column sums equal to12 and 30.
After some trial and error: If t=1 and y=3 then y*(1 + t + t^2 + t^3) = 3*4 =12 If x=z=2 then x*(1+ z + z^2 +z^3) = 2*15 =30.
So x=2 y=3 z=2 t=1.seems to check out for all 4 equations. I know this was guess work but as syber said "It's the journey that counts!"

[TheShacharZiv]

I subtracted and got x(z-1)+ y(t-1)= 2. Assuming positive integers x cannot be 1. If x=2, then z=2, t=1, and y=3.

[StaR-uwdc]

nice solution, in the end it turned out that there was no need to multiply the first equation by z+t

[pwmiles]

The RHS differences are 2, 4, 8. This suggests a geometric progression with base 2. With a little experimentation I got
xz^0 + yt^0 = 3 + 2.2^0
xz^1 + yt^1 = 3 + 2.2^1
xz^2 + yt^2 = 3 + 2.2^2
xz^3 + yt^3 = 3 + 2.2^3

A solution is x=3, z=1, y=2, t=2

(x, y) and (z, t) can be exchanged. Other roots are not ruled out though.

[Qermaq]

I'm laughing because you went through the trouble of putting the givens on each page but continued to just scroll up to the top every time you wanted them! I guess practice is in order ;)