In a book on mathematical induction, I found the following problem:
Let $p$ be a prime number of the form $4k+3$. Prove that for all $n \in \mathbb{N}$,
$\displaystyle\sum\limits_{i=1}^{2 k+1} i^{2^{n}}$
is divisible by $p$.
So I applied this in the particular case where $11=4\cdot2+3$
Prove that for all $n\in\mathbb{N},$
$11 \mid 1+2^{2^n}+3^{2^n}+4^{2^n}+5^{2^n}.$
So, I wrote $P(k+1)$ as follows: $1+(2^{2^n})^2+(3^{2^n})^2+(4^{2^n})^2+(5^ {2^n})^2$
Now, I'm stuck, I can't apply my inductive hypothesis in such a way that proves $P(k+1)$. What do you think of this discussion?
For the particular case $p=11$, define $P(n): 11 | 1+2^{2^n}+3^{2^n}+4^{2^n}+5^{2^n} $. $P(1)$ holds, and suppose $P(k)$ with $k>1$ holds too. Then we need to show $P(k+1) \implies 11 | 1+4^{2^k}+9^{2^k}+16^{2^k}+25^{2^k}$ holds. But in the ring of integers $Z_{11}$, $9 = -2$, $16=5$ and $25=3$, since the characteristic is $11$. Therefore, $P(k+1)$ is the same thing as saying $11| 1 + 4^{2^k}+(-2)^{2^k}+5^{2^k}+3^{2^k}$, which is very certainly true by our inductive hypothesis $P(k)$.