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• It all started when I noticed the following: 4*6 = 24 5*5 = 25 6*4 = 24 And that 24 was very close to 25. I wondered whether if I have a square a*a I can
Message 1 of 1 , Dec 10 9:32 AM
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It all started when I noticed the following:

4*6 = 24
5*5 = 25
6*4 = 24

And that 24 was very close to 25. I wondered whether if I have a square a*a I
can subtract 1 from one multiplicand and add 1 to the other and get the same
result. Let's see:

(a-1)*(a+1) = a^2-a+a-1=a^2-1

Nope! However, now that I tought about it I realised that it will hold for
any "a" no matter how large or how small, which is nice. So 999*1001 =
1000^2-1.

Now let's generalise our requirements a bit: what if we have a product of "a"
times "b", and we want to do the same:

(a-1)*(b+1) = ab
ab-b+a-1 = ab
a - b = 1

So if we increase a number by one and decrease the other by one, we'll get the
same product only if their difference is 1. I.e: we switched them.

But what if we decrease and increase by a \delta (LaTeX notation) that is not
necessarily 1?

(a-\delta)*(b+\delta) = ab
ab-\delta(b-a) + \delta^2 = ab
\delta^2 = \delta(a-b)
a-b = \delta

So again there's no solution except the trivial one.

I wonder if there are non-trivial solutions for (a-1)b(c+1) or (a-1)(b+1)(c-1)
(d+1) or any other products like that, but that would probably be somewhat
hairier algebra.

Regards,

Shlomi Fish

---------------------------------------------------------------------
Shlomi Fish shlomif@...
Homepage: http://www.shlomifish.org/

Chuck Norris wrote a complete Perl 6 implementation in a day but then
destroyed all evidence with his bare hands, so no one will know his secrets.
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