Let the fractions be f1 = a/b and f2 = c/d, a,b,c,d all positive integers
O1 = f1 + f2 = a/b + c/d = (ad)/bd) + (cb)/(bd) = (ad + bc)/(bd)
O2 = f1 * f2 = (a/b) * (c/d) = (ac)/(bd)
Since O1 = O2, then (ad + bc)/(bd) = (ac)/(bd)
i.e. ad + bc = ac
Assuming f1 and f2 are proper fractions, then we can write,
b = a + ε, d = c + η, ε, η positive integers
Substitution then gives us,
a(c + η) + (a + ε)c = ac
ac + aη + ac + εc = ac
ac + aη + εc = 0
but a,c,η,ε are all positive, which makes the last equation invalid.
Therefore the fractions, f1 and f2 must be improper fractions with
b = a – ε, d = c – η, ε, η positive integers
Substitution then gives us,
a(c – η) + (a – ε)c = ac
ac – aη + ac – εc = ac
ac – aη – εc = 0
a(c – η) = εc
Now it is just a matter of choosing (guessing at) appropriate values for the unknown integers in the last equation.
Taking c = 7 and η = 2, then d = 7 – 2 = 5, giving a(c – η) = εc as 5a = 7ε
The solution to 5a – 7ε = 0 is a = 7t, ε = 5t, t = 1,2,3,...
And b = a – ε = 7t – 5t = 2t
Giving a/b = 7t/2t = 7/2
And c/d = 7/5
Answer: two suitable fractions are: f1 = 7/2 and f2 = 7/5
Check
O1 = f1 + f2 = a/b + c/d = 7/2 + 7/5 = 35/10 + 14/10 = 49/10
O2 = f1 * f2 = (a/b) * (c/d) = (7/2) *(7/5) = 49/10
i.e. O1 = O2