Solving a Diophantine System

This video is about solving a Diophantine System
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#NumberTheoryProblems #DiophantineEquations

EXPLORE:

[Qermaq]

Nice and not so easy for me. I could figure out the answers easily enough in my head, but to prove it? That was the feat. You missed an ordering of the 1, 5, 2, 3 options, replacing it with the already determined 2, 2, 2, 2. Other than that spot on.

[fredericoapuleio]

I love this problem. Before I saw your solution I did it this way: rewriting the equations as x + y = zw = P, xy = z+w= Q. So we can say that x and y are solutions of t2 - Pt + Q=0, z and w are solutions of t2-Qt+P=0. In order to exist solutions we must have P2-4Q=m2 and Q2-4P=n2. This way we can parametrize the solutions and generate not only the integer solutions but any other.

[elkincampos]

x+y=z*w and z+w=x*y. Then x, y are roots the polynomial f(t)=t^2-(z*w)*t+z+w. But are roots of f(t) are roots iff (z*w)^2-4*(z+w) is a square. Therefore (z*w)^2-4*(z+w)<=(z*w-2)^2, Why?(Mod 4). Thus -4*(z+w)<=-4*w*z+4. That is w*z<=z+w+1. We suppose that w<=z. Now if w>=3 then w*z>=3*z=z+z+z>z+w+1 contradiction. It follows w=1 or 2. If w=2 then w*z=2*z=z+z<=z+w+1. Then z=2 or 3.

If w=1 then w*z=z<z+w.

Now if w=2 then z=2 or 3. . It follows that z^2-z-2 is a square and z=3 or 2 si z=2 then z^2-z-2 =0 and 4*(z^2-z-1)=4=0. If z=3 then z^2-z-2=4=2^2 .
If w=1 then Then It follows z<=5. If z=2 then z^2-4*z-4=4-8-4<0 . If z=1 then z^2-4*z-4<0. Contradiction. If z=3 then z^2-4*z-4<0. if z=4 then z^2-4*z-4<0. If z=5 then z^2-4*z-4=25-20-4=1.

We conclude z=5, w=1. And z=2, w=2. z=2, w=3.

[angelmendez-rivera]

Once the equation (x – 1)·(y — 1) + (z – 1)·(w – 1) = 2, one can intuitively proceed by using the restriction. At the beginning of the video, a specific restriction was stated: that x >= 1, y >= 1, z >= 1, w >= 1, aside from the x, y, z, w being integers. Therefore, the smallest possible value for either (x – 1)·(y — 1) or (z – 1)·(w – 1) is 0. This in turn implies the biggest possible value of the other is 2. This is how you can intuitively conclude that the sums must 0 + 2 = 2, 1 + 1 = 2, 2 + 0 = 2. This gives three cases, each with a separate system of equations: (a) (x – 1)·(y – 1) = 0 & (z – 1)·(w – 1) = 2 (b) (x – 1)·(y – 1) = 1 & (z – 1)·(w – 1) = 1 (c) (x – 1)·(y – 1) = 2 & (z – 1)·(w – 1) = 0.

[brsram]

One more solution is there, (1, 5, 2, 3)

[ashishpradhan]

Today I have my maths olympiad. It's geometry portion went preety good but not the number theory one.
I have also one diophantine equation . Could you verify it's answer?