rmo aakash lecture 11 pending 21 oct 2026 - functions - Q1 done

Q1.

S1.

f(2) = 3
f(2^2) = f(4) = 3.3 = 9
2 < 3 < 4
=> f(2) < f(3) < f(4)
=> 3 < f(3) < 9
=> f(3) can be 4,5,6,7,8
Let f(3) = a.

Now,
f(3^2) = f(9) = f(3).f(3)
8 < 9 < 16
=> f(8) < f(9) < f(16)
=> 27 < f(9) = a^2 < 81
=> 27 < a^2 < 81
=> a is 6,7,8 (Since 4,5 doesn't work)

Now
16 < 27 < 32
f(16) < f(27) < f(32)
81 < a^3 < 243
=> a = 6 (7,8 don't work)

So f(3) = 6

Now,
Powers of 2: 2,4,8,16,32,64,128,256
Powers of 3: 3,9,27,81,243
f(243) < f(256)
6^5 < 3^8
3^5.2^5 < 3^8
2^5 < 3^3
32 < 27
contradiction.
H.P.


Q2.

S2.
Let S be the set of all values of f(n) for n >= 1.
From well ordering principle, there will be a minimum value in this set(made up of non zero integers).
Let that value be d.
Let that value occur at m => f(m) = d

Now, there are 2 cases, m = 1 or m >= 1.
If m >= 1, we can write the given inequality as:
f(m) > f(f(m-1)) If f(m-1) = t then f(m) > f(t) => d > f(t) => d is not the minimum. contradiction.
So m has to be 1.
So f(1) has to be minimum in set S.

Similarly if S' is made up of all f(n) for n >= 2 then similarly we can show that f(2) will be the minimum in this S'.

=> f(1) < f(2) < f(3) ...
Since f(1) >=1 => f(2) >= 2, f(3) >= 3 and so on...

Now, let's say there is some k for which f(k) > k, then f(k) >= k + 1
Earlier we showed that the function is increasing.
So
f(f(k)) >= f(k + 1)
which contradicts the original question statement.
So for no 'k', f(k) > k.
=> f(k) = k
H.P.

Q3. Find all functions f: N -> N s.t. 
f(f(m) + f(n)) = m + n for all m,n belong to N.
S3.
trivial solution f(x) = x.
Also, it can't be constant function clearly.