Can a Googol Be Written as n^2-m^2?

[anemone]

Hi MHB,

It's me, anemone.

I saw this problem quite some time ago, and I don't remember where...but it says "Can a googol $10^{100}$ be written as $n^2-m^2$, where $n$ and $m$ are positive integers?"


Since I don't know where I should start to think about how to answer it, I thought there would be no harm in asking about it here because MHB knows it all!(Sun)

Thanks in advance for any input that anyone is going to give me.:)

 

[I like Serena]

Hi MHB,

It's me, anemone.

I saw this problem quite some time ago, and I don't remember where...but it says "Can a googol $10^{100}$ be written as $n^2-m^2$, where $n$ and $m$ are positive integers?"


Since I don't know where I should start to think about how to answer it, I thought there would be no harm in asking about it here because MHB knows it all!(Sun)

Thanks in advance for any input that anyone is going to give me.:)

Let's take a look at factorization.
\begin{array}{lcl}n^2-m^2 &=& 10^{100}\\
(n+m)(n-m) &=& 2^{100} 5^{100}\end{array}

Suppose we pick \(\displaystyle n+m = 2^{50} 5^{51}\) and \(\displaystyle n-m=2^{50} 5^{49}\).
Then the solution
\begin{cases}n&=&10^{49}(25+1)\\
m&=&10^{49}(25-1)\end{cases}
comes rolling out...

There are bound to be more solutions.

 

[anemone]

Let's take a look at factorization.
\begin{array}{lcl}n^2-m^2 &=& 10^{100}\\
(n+m)(n-m) &=& 2^{100} 5^{100}\end{array}

Suppose we pick \(\displaystyle n+m = 2^{50} 5^{51}\) and \(\displaystyle n-m=2^{50} 5^{49}\).
Then the solution
\begin{cases}n&=&10^{49}(25+1)\\
m&=&10^{49}(25-1)\end{cases}
comes rolling out...

There are bound to be more solutions.

Hi I like Serena,

Thank you so much for the reply!

Now I know of a better concept whenever I want to solve any math problems..i.e. to play with the exponents! (Sun)

I appreciate it you made the explanation so clear for me! Thanks!

 

[soroban]

Hello, anemone!

$$\text{Can a googol }10^{100}\text{ be written as }n^2-m^2\text{, where }n\text{ and }m\text{ are positive integers?}$$

Any multiple of 4 can be written as a difference of squares.Example: Express 80 as a difference of squares.

A multiple of 4 can be expressed as the sum of two consecutive odd integers.
. . We have: .$$80 \:=\:39+41$$

Consecutive squares differ by consecutive odd integers.

We have: .$$\begin{array}{ccccc}19^2 && 20^2 && 21^2 \\ \hline & 39 && 41 \end{array}$$

Therefore: .$$80 \;=\;21^2 - 19^2$$
Let $$G = 10^{100}.$$

The two odd numbers are: .$$\tfrac{G}{2}-1\text{ and }\tfrac{G}{2}+1$$

The two squares are: .$$\begin{Bmatrix}\frac{(\frac{G}{2}-1)-1}{2} &=& \tfrac{G}{4}-1 \\ \frac{(\frac{G}{2}+1)+1}{2} &=& \tfrac{G}{4}+1 \end{Bmatrix}$$Therefore: .$$G \;=\;\left(\tfrac{G}{4}+1\right)^2 - \left(\tfrac{G}{4} - 1\right)^2$$

 

[anemone]

Hello, anemone!


Any multiple of 4 can be written as a difference of squares.Example: Express 80 as a difference of squares.

A multiple of 4 can be expressed as the sum of two consecutive odd integers.
. . We have: .$$80 \:=\:39+41$$

Consecutive squares differ by consecutive odd integers.

We have: .$$\begin{array}{ccccc}19^2 && 20^2 && 21^2 \\ \hline & 39 && 41 \end{array}$$

Therefore: .$$80 \;=\;21^2 - 19^2$$
Let $$G = 10^{100}.$$

The two odd numbers are: .$$\tfrac{G}{2}-1\text{ and }\tfrac{G}{2}+1$$

The two squares are: .$$\begin{Bmatrix}\frac{(\frac{G}{2}-1)-1}{2} &=& \tfrac{G}{4}-1 \\ \frac{(\frac{G}{2}+1)+1}{2} &=& \tfrac{G}{4}+1 \end{Bmatrix}$$Therefore: .$$G \;=\;\left(\tfrac{G}{4}+1\right)^2 - \left(\tfrac{G}{4} - 1\right)^2$$

Thank you so much soroban for letting me know of this useful and handy knowledge...I really appreciate it!:)