Let us write $ LH= f^{19}+ 97 f(n) = 98 n + 232 = RH$ One notes that: 1°. The function $ f: N \rightarrow N$ is strictly increasing. 2°. $ f^{19}+ 97 f(n)$ needs to be even number, since RH is always even. Let $ n \longmapsto k\cdot n$ , so $ f(n) = k\cdot n, k \in N$ , then $ f^{19}(n) = k^{19} \cdot n$, $ LH= k^{19}n + 97 k\cdot n = 98 n + 232$. For $ n=1$, we have $ k^{19} + 97 k = 330$, so it is not a solution. Let $ f(n) = n + b, b \in N$, then $ f^{2}(n)= n+ 2 b$, $ f^{19}(n)= n+ 19 b$. So we have $ n + 19 b + 97 ( n+b) - 98 n = 232$ Therefore, $ (19 + 97) b = 232$ so $ 116 b = 232$, and $ b = 2$. We have $ n+ 38 + 97 ( n+2) - 98 n = 232$ $ 38 + 194 = 232$, hence $ f(n) = n + 2$ that is a possible solution. $ f^{19}(n) +97 f(n)= 98n +232 =RH$ RH is a 1st degree polynomial. Hence $ LH= f^{19}(n) +97 f(n)$ must be 1st degree polynomial.That is only possible if $ LH=f(n)= a n+b$ But if $ a\in N$ then $ f^{19}(n)=a^{19}n+(a^{18}+...+a+1)b$. So $ a^{19}n+(a^{18}+...+a+1)b+97(an+b)-98n=232$, $ (a^{19}+97a-98)n=232-(a^{18}+...+a+1)b-97b$. So, $ (a^{19}+97a-98)=0$ only solution is for $ a=1$.

Suppose $ f(n)^{19} + 97f(n) = 98n + 232$ holds. Picking $ n = 1$ we see that $ f(1)^{19} + 97f(1) = 330 \Rightarrow f(1) [ f(1)^{18} + 97] = 330$ We conclude that $ f(1)$ is a divisor of 330. It is easy to see that for $ f(1) = 1$ equality doesn't hold and, if $ f(1) > 1$ then $ f(1)^{18} \ge 330$. So, no such $ f$ exists!

azo wrote: Suppose $ f(n)^{19} + 97f(n) = 98n + 232$ holds. Picking $ n = 1$ we see that $ f(1)^{19} + 97f(1) = 330 \Rightarrow f(1) [ f(1)^{18} + 97] = 330$ We conclude that $ f(1)$ is a divisor of 330. It is easy to see that for $ f(1) = 1$ equality doesn't hold and, if $ f(1) > 1$ then $ f(1)^{18} \ge 330$. So, no such $ f$ exists! I think $ f(n)^{19}=f(f(...(f(n)...))$ nineteen times. At least that's a common notation, I think. Also, if it meant $ f(n)^{19}$, that would make the problem trivial as you noticed, so I think my interpretation is correct

Kabor Suk wrote: Let us write $ LH = f^{19} + 97 f(n) = 98 n + 232 = RH$ One notes that: 1°. The function $ f: N \rightarrow N$ is strictly increasing. 2°. $ f^{19} + 97 f(n)$ needs to be even number, since RH is always even. Let $ n \longmapsto k\cdot n$ , so $ f(n) = k\cdot n, k \in N$ , then $ f^{19}(n) = k^{19} \cdot n$, $ LH = k^{19}n + 97 k\cdot n = 98 n + 232$. For $ n = 1$, we have $ k^{19} + 97 k = 330$, so it is not a solution. Let $ f(n) = n + b, b \in N$, then $ f^{2}(n) = n + 2 b$, $ f^{19}(n) = n + 19 b$. So we have $ n + 19 b + 97 ( n + b) - 98 n = 232$ Therefore, $ (19 + 97) b = 232$ so $ 116 b = 232$, and $ b = 2$. We have $ n + 38 + 97 ( n + 2) - 98 n = 232$ $ 38 + 194 = 232$, hence $ f(n) = n + 2$ that is a possible solution. $ f^{19}(n) + 97 f(n) = 98n + 232 = RH$ RH is a 1st degree polynomial. Hence $ LH = f^{19}(n) + 97 f(n)$ must be 1st degree polynomial.That is only possible if $ LH = f(n) = a n + b$ But if $ a\in N$ then $ f^{19}(n) = a^{19}n + (a^{18} + ... + a + 1)b$. So $ a^{19}n + (a^{18} + ... + a + 1)b + 97(an + b) - 98n = 232$, $ (a^{19} + 97a - 98)n = 232 - (a^{18} + ... + a + 1)b - 97b$. So, $ (a^{19} + 97a - 98) = 0$ only solution is for $ a = 1$. f(n) need not to be a polynomial and that the difficulty of this problem

I believe $ f^{(19)}$ means composition, not power. Otherwise azo already shows it is trivial. We have $ f(n)< \frac{98n+232}{97}=n+2+\frac{n+38}{97}$. So $ f(n)\le n+2$ when $ n\le59$. It is not hard to show, $ f^{(19)}(n)\le n+38,\forall n\le21$ Then for $ n\le21$, $ 98n+232\le97(n+2)+n+38\Rightarrow f(n)=n+2,f^{(19)}(n)=n+38$. Actually $ f^{(19)}(20)=20+38$ and $ f^{(19)}(21)=21+38$ imply $ f(n)=n+2,\forall n\le57$. Now we will prove $ f(n)=n+2,\forall n$ by induction. Suppose $ f(n)=n+2,\forall n<m$. We have $ f^{(19)}(m-36)=f(m)=98*(m-36)+232-97*f(m-36)\\=98*(m-36)+232-97*(m-34)=m+2$. Q.E.D.

xxp2000 wrote: It is not hard to show $f^{(19)}(n)\le n+38,\forall n\le21 $ It is not obvious for me. please show it.

xxp2000 wrote: So $ f(n)\le n+2$ when $ n\le59$. It is a silly question to ask to someone who made his post two years ago ... However, then $ f(f(n))\le f(n)+2\le n+4$ when $ n\le57$ etc. (step by step, until) $ f^{(19)}(n)\le f^{(18)}(n) + 2 \le n+38$ when $n\le23$ (even more than the $n\le 21$ used).

An easy solution, first checked each side$\mod 97$ then $f^{19}(x) \equiv n+38 \mod 97 \implies $ $\forall n\le 58 : f^{19}(x) \ge n+38 \implies f^{19}(x)+97f(n) =98n+232 \ge n+38 + 97f(n) \implies n+2 \ge f(n) $ and now it will imply $ n+38 \ge ...\ge f^{17}(x)+4\ge f^{18}(x)+2\ge f^{19}(x)$ and thereupon because $f^{19}(x) \ge n+38, f^{19}(x) \le n+38 \implies f^{19}(x)=n+38 \implies f^{19}(x)+97f(n)=n+38+97f(n)=98n+232$ $\implies \forall n \le 58 : f(n)=n+2 $ and the rest is straightforward using induction. suppose we've proven that for $n>36$ , $\forall i < n, : f(i)=i+2 \implies f^{19}(n-36)+f(n-36)=98(n-36)+232$ but from induction hypothesis we have: $f(n-36)=n-34$ and $f^{19}(n-36)=f^{18}(n-34)=f^{17}(n-32)=...=f(n)$ and now substituting the values we see: $f^{19}(n-36)+f(n-36)=f(n)+97(n-34)=98(n-36)+232 \implies f(n)=n+2$ and induction step in complete