I need assistance to prove with the principle of mathematical induction that for every $n \in \Bbb Z^+$, there exists natural numbers $a_n$ and $b_n$ such that $5^n={a_n}^2+{b_n}^2$.
So far, I've managed to prove it for $P_1$ and $P_2$, but am unsure on what to do to continue in the inductive step.
Thanks again!!!
On
Consider,
$$5^{1}=1^2+2^2$$
$$5^{2}=3^2+4^2$$
$$5^{3}=5^{2}(1^2+2^2)$$
$$5^{4}=5^{2}(3^2+4^2)$$
$5^{n+2}=5^2(5^n)$. If $5^n=a_n^2+b_n^2$ then $5^{n+2}=(5a_{n})^2+(5b_{n})^2$. So if $5^n$ can be expressed as the sum of two positive squares for $n=1$ then it can be expressed as the sum of two positive squares for $n=1+2=3$ and hence $n=5$ etc (for all positive odd $n$). If $5^n$ can be expressed as the sum of two positive squares for $n=2$, then it can be expressed as the sum of two positive squares for $n=4$ and hence for $n=6$ etc (for all positive even $n$).
Define $P_n=5^n=a_n^2+b_n^2$
For $P_1=1^2+2^2$ and $P_2=3^2+4^2$, the criteria holds true. And so you're almost done.
Note that if $n=2k$ and $P_{n+2}=5^2 \cdot 5^n = 5^2(a_n^2+b_n^2)=(5a_n)^2+(5b_n)^2$. And you get the $P_{\text {evens}}$ this way. Like, $P_2=3^2+4^2$. So, $P_{2+2=4}=5^2(3^2+4^2)=15^2+20^2$ and like that you get the evens.
For $n=2k+1$, $P_{n+2}=(5a_n)^2+(5b_n)^2$ in the similar manner to get the $P_{\text{odds}}$.
Induction over!
And I chose to separate out the odds and evens because $P_{\text {odd}}=5^{\text {odd}}=a_\text {odd}^2+b_{\text {odd}}^2$. From that, only if you multiply it with $5^2$, you get RHS to be in the form of two squares right? Think in a similar manner for the even's case.
And basically, if I had taken something like $5^n=a_n^2+b_n^2$ and tried to multiply it with a $5$ to get $5^{n+1}$, then RHS would have been like $5a_n^2+5b_n^2$ and that's no way the sum of two perfect squares. We definitely have a perfect square. But by induction it can't be shown in that case.
I hope everything is clear now. More clear perhaps!