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

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A googol, represented as 10^100, can indeed be expressed as the difference of two squares, n^2 - m^2, where n and m are positive integers. The factorization approach shows that n^2 - m^2 equals (n+m)(n-m), which can be set to the prime factorization of 10^100, specifically 2^100 * 5^100. By selecting specific values for n+m and n-m, valid solutions for n and m can be derived, confirming the possibility of expressing a googol in this form. Additionally, it is noted that any multiple of 4 can be represented as a difference of squares, further supporting the argument. The discussion highlights the mathematical principles involved in solving such problems.
anemone
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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.:)


 
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anemone said:
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 $$n+m = 2^{50} 5^{51}$$ and $$n-m=2^{50} 5^{49}$$.
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.
 
Last edited:
I like Serena said:
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 $$n+m = 2^{50} 5^{51}$$ and $$n-m=2^{50} 5^{49}$$.
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!
 
Last edited by a moderator:
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
 
soroban said:
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!:)
 
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