A number theory problem by Shwet Ranjan

• Fermat's theorem says that a^p = a(mod p) where p is a prime and a is an integer.

Since $\frac{a^p}{p}$ - $\frac{a}{p}$ is an integer.

$\frac{K^7}{7}$ + $\frac{K^5}{5}$ + $\frac{2K^3}{3}$ - $\frac{k}{105}$

= ( $\frac{k^7}{7}$ - $\frac{k}{7}$ ) + ( $\frac{k^5}{k}$ - $\frac{k}{5}$ )+2( $\frac{k^3}{3}$ - $\frac{k}{3}$ ) + $\frac{k}{7}$ + $\frac{k}{5}$ + $\frac{2k}{3}$ - $\frac{k}{105}$

= ( $\frac{k^7}{7}$ - $\frac{k}{7}$ ) + ( $\frac{k^5}{k}$ - $\frac{k}{5}$ )+2( $\frac{k^3}{3}$ - $\frac{k}{3}$ ) + k

is an integer.

Let $f(K) = \dfrac{K^7}{7} + \dfrac{K^5}{5}+\dfrac{2K^3}{3}-\dfrac{K}{105}$ .

When $K = 1$ ,

$f(1) = \dfrac{1}{7} + \dfrac{1}{5}+\dfrac{2}{3}-\dfrac{1}{105} = 1$ which is an integer.

Assume that when $K = k, f(k)$ is an integer.

Now when $K = k + 1$ ,

$f(k + 1) = \dfrac{(k + 1)^7}{7} + \dfrac{(k + 1)^5}{5}+\dfrac{2(k + 1)^3}{3}-\dfrac{(k + 1)}{105}$

$f(k + 1) = \dfrac{k^7}{7} + \dfrac{k^5}{5} + \dfrac{2k^3}{3} - \dfrac{k}{105} + \dfrac{1}{7} + \dfrac{1}{5} + \dfrac{2}{3} - \dfrac{1}{105} + \dfrac{\binom{1}{7}k^6 + \cdots + \binom{6}{7}k}{7} + \dfrac{\binom{1}{5}k^4 + \cdots + \binom{4}{5}k}{5}+2\times\dfrac{\binom{1}{3}k^2 + \binom{2}{3}k}{3}$

And since $_7C^n | 7$ where $0< n <7$ , $_5C^s | 5$ where $0< s <5$ and $_3C^g | 3$ where $0< g <3$ and we previously stated that $f(1)$ are $f(k)$ are integers, it follows that $f(k + 1)$ is also an integer $\therefore f(K)$ is an integer $\forall K$ and the answer is $\boxed{\text{True}}$ .