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A googol

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anemone

MHB POTW Director
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Feb 14, 2012
3,684
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.:)


 

Klaas van Aarsen

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Mar 5, 2012
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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.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,684
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!
 
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soroban

Well-known member
Feb 2, 2012
409
Hello, anemone!

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

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: .[tex]80 \:=\:39+41[/tex]

Consecutive squares differ by consecutive odd integers.

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

Therefore: .[tex]80 \;=\;21^2 - 19^2[/tex]



Let [tex]G = 10^{100}.[/tex]

The two odd numbers are: .[tex]\tfrac{G}{2}-1\text{ and }\tfrac{G}{2}+1[/tex]

The two squares are: .[tex]\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}[/tex]


Therefore: .[tex]G \;=\;\left(\tfrac{G}{4}+1\right)^2 - \left(\tfrac{G}{4} - 1\right)^2[/tex]
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,684
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: .[tex]80 \:=\:39+41[/tex]

Consecutive squares differ by consecutive odd integers.

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

Therefore: .[tex]80 \;=\;21^2 - 19^2[/tex]



Let [tex]G = 10^{100}.[/tex]

The two odd numbers are: .[tex]\tfrac{G}{2}-1\text{ and }\tfrac{G}{2}+1[/tex]

The two squares are: .[tex]\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}[/tex]


Therefore: .[tex]G \;=\;\left(\tfrac{G}{4}+1\right)^2 - \left(\tfrac{G}{4} - 1\right)^2[/tex]
Thank you so much soroban for letting me know of this useful and handy knowledge...I really appreciate it!:)