# Thread: lovely result, but curious

1. ## lovely result, but curious

Hi,

Whilst meandering through integers that fit the equation

$\displaystyle a^2+b^2=2c^2$

I came across this curious result

$\displaystyle 1^2+7^2=2(5^2)$
$\displaystyle 11^2+77^2=2(55^2)$
$\displaystyle 111^2+777^2=2(555^2)$

I went as far as 16 digits and it still holds true.

$\displaystyle 1111111111111111^2+7777777777777777^2=2(5555555555 555555^2)$

I stumped when it comes to figuring a reason for the repetion though.

Any ideas anyone?

Thanks

Pro

2. ## Re: lovely result, but curious

$\displaystyle 1^2+7^2=2(5)^2$

$\displaystyle 11^2\left(1^2+7^2\right)=2(11)^2\;5^2\Rightarrow\ 11^2+\left[11(7)\right]^2=2\left[11(5)\right]^2$

You can continue with

$\displaystyle 111^2\left(1^2+7^2\right)=2(111)^2\;5^2$

and so on.

3. ## Re: lovely result, but curious

Thanks Archie,

I knew there must be an answer, I just couldn't see it for looking, lol

Pro

4. ## Re: lovely result, but curious

Hello, procyon!

Whilst meandering through integers that fit the equation $\displaystyle a^2+b^2\:=\:2c^2$

I came across this curious result
. . $\displaystyle \begin{array}{ccc}1^2+7^2 &=& 2(5^2) \\ 11^2+77^2 &=& 2(55^2) \\ 111^2+777^2 &=& 2(555^2) \\ \vdots && \vdots \end{array}$

I went as far as 16 digits and it still holds true.
I stumped when it comes to figuring a reason for the pattern though.
Any ideas anyone?

$\displaystyle \begin{array}{cc}\text{An integer consisting of }n\text{ 1's has the form:} & \dfrac{10^n-1}{9} \\ \\[-3mm] \text{An integer consisting of }n\text{ 7's has the form:} & 7\!\cdot\!\dfrac{10^n-1}{9} \\ \\[-3mm] \text{An integer consisting of }n\text{ 5's has the form:} & 5\!\cdot\!\dfrac{10^n-1}{9} \end{array}$

We wish to show that: .$\displaystyle \left(\frac{10^n-1}{9}\right)^2 + \left(7\!\cdot\!\frac{10^n-1}{9}\right)^2 \;=\;2\left(5\!\cdot\!\frac{10^n-1}{9}\right)^2$

The left side is: .$\displaystyle \left(\frac{10^n-1}{9}\right)^2 + 49\left(\frac{10^n-1}{9}\right)^2 \;=\;50\left(\frac{10^n-2}{9}\right)^2$

. . . . . . . . . . $\displaystyle =\;2\cdot25\left(\frac{10^n-1}{9}\right)^2 \;=\; 2\left(5\!\cdot\!\frac{10^n-1}{9}\right)^2$
Q.E.D.

5. ## Re: lovely result, but curious

Hi Soroban,

Very nice proof

Thank you

Pro

6. ## Re: lovely result, but curious

Originally Posted by procyon
Thanks Archie,

I knew there must be an answer, I just couldn't see it for looking, lol

Pro
All you have is "two sides are equal, so any common multiple of those two sides are equal"

$\displaystyle 1^2+7^2=(2)5^2$

$\displaystyle k^2\left(1^2+7^2\right)=2k^2\left(5^2\right)$

In this case, $\displaystyle k^2=\left[(1.1)10^n\right]^2$, for n=1, 2, 3,.....