Proving that $5^n$ can be written as the sum of two squares, by induction

Question

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!!!

As an aside, this follows immediately from the sum of two squares theorem, but I imagine you want to prove this without such heavy machinery from first principles. — JMoravitz, Mar 24 '17 at 03:48
Yes please, I'm only a year 12 student, so heavy machinery or university grade concepts would probably blow me away! — Maths Matador, Mar 24 '17 at 03:51

score 5 · Answer 1 · answered Mar 24 '17 at 03:50

5

Hint: Consider $(2a-b)^2 + (2b+a)^2$.

answered Mar 24 '17 at 03:50

Thomas Andrews

177,126

@W.Nguyen in case this hint is still confusing and since a full answer has been added elsewhere using a different method, to explain this one note that $(2a-b)^2+(2b+a)^2=5(a^2+b^2)$, so if $5^n = a_n^2+b_n^2$ one has $5^{n+1}=5(a_n^2+b_n^2)=(2a_n-b_n)^2+(2b_n+a_n)^2$ is the sum of two squares as well. (in order to avoid possible negative numbers, we may assume without loss of generality that $a_n$ is larger than $b_n$). This answer avoids the need of running induction on odd and even numbers separately and instead handles it all at once. – JMoravitz Mar 24 '17 at 04:25

Mathejunior · Accepted Answer · 2017-03-24T09:25:19.910

1

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!

edited Mar 24 '17 at 09:25

answered Mar 24 '17 at 03:56

Mathejunior

3,344

I'm sorry but I don't follow what you've done for $n=2k$ and everything after it. Could you guide me through the process you took please? As in, why did you use $2k$ and what does it really mean? – Maths Matador Mar 24 '17 at 04:01
Did you understand now? – Mathejunior Mar 24 '17 at 04:04
Yes! That helps a lot, thank you!! – Maths Matador Mar 24 '17 at 04:05
@W.Nguyen I have added my approach clearly. – Mathejunior Mar 24 '17 at 04:22

Ahmed S. Attaalla · Answer 3 · 2017-03-24T14:35:56.460

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 $n$).

Proving that $5^n$ can be written as the sum of two squares, by induction

3 Answers3