gistfile1.txt

Theorem.
There exist no 2 pairs of integers {a,b},{c,d} such that ab = cd(#1) ,a+b = c+d(#2), {a,b} != {c,d}, and a,b,c,d > 2

Proof.
Suppose we have {a,b},{c,d} that satisfies all the conditions above.

We can suppose a,b,c,d are not all equal without losing generality,
since if a = b, a^2 = cd (from #1) and 2a = c+d (from #2)
=> 4a^2 = (c+d)^2
=> 1/4(c+d)^2 = cd
=> (c+d)^2 = 4cd
=> c^2 + 2cd + d^2 = 4cd
=> (c-d)^2 = 0
=> c = d
but if a=b and c=d, ab = cd implies a^2 = c^2 and thus a = c, therefore a,b,c,d are all equal.
Which contradicts the condition that {a,b} != {c,d}

furthermore, since a,b,c,d are not all equal, we can suppose a < b, c < d, and d > b without losing generality.
Therefore c < a < b < d.

From #2, we have a-c = d-b.
Let r = a-c = d-b,  
=> r > 0, a = c+r, d = b+r
since ab = cd, (c+r)b = c(b+r)
=> cb+br = cb+cr => br = cr => b = c
Which contradicts the condition that a,b,c,d are not all equal. Q.E.D.