# A googol

#### anemone

##### MHB POTW Director
Staff member
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! Thanks in advance for any input that anyone is going to give me. #### Klaas van Aarsen

##### MHB Seeker
Staff member
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! 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.

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#### anemone

##### MHB POTW Director
Staff member
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! I appreciate it you made the explanation so clear for me! Thanks!

Last edited by a moderator:

#### soroban

##### Well-known member
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

##### MHB POTW Director
Staff member
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! 