Prove that when all expessions of the form:
+/- √1 +/- √2 +/- ....... +/- √100 are multiplied together, the result is an integer.
As an example, multiplying all expressions of the form: +/- √1 +/- √2 is equivalent to finding the result of this product:
(√1 +√2)( √1 - √2)( - √1 + √2)( - √1 - √2)

**** Extra Challenge: Solve this puzzle without the aid of a computer program.

Establish the proposition for (a+b)(a-b)(-a+b)(-a-b) = (a^2-b^2)^2, with each element occurring to an even power, and a single difference of squares.

Let (a+b) = k, then -a-b=-k; let (a-b)=l; then (-a+b) = -l [1]

(k+c)(k-c)(l+c)(l-c)(-k+c)(-k-c)(-l+c)(-l-c) = (c^2-k^2)^2 (c^2-l^2)^2 with each element again occurring to an even power, and two differences of squares. Since all we have done with each of k and l is just the same as we did before with a and b, each of the latter on resubstitution must equally occur to an even power; in fact the expression is equivalent to:

a^8-4a^6b^2-4a^6c^2+6a^4b^4+4a^4b^2c^2+6a^4c^4-4a^2b^6+4a^2b^4c^2+4a^2b^2c^4-4a^2c^6+b^8-4b^6c^2+6b^4c^4-4b^2c^6+c^8

In the same way as [1], we can pair off (k+c)(k-c)(-k+c)(-k-c) to m,-m,n,-n, and (l+c)(l-c)(-l+c)(-l-c) to o,-o,p,-p and repeat the whole thing for d, giving (d^2-m^2)^2 (d^2-n^2)^2(d^2-o^2)^2 (d^2-p^2)^2, and so on...

So the nth reiteration of the same exercise will be the product of 2n squared differences of squares, producing an integer result, as was to be shown.

Edited on September 19, 2015, 12:53 am

Posted by broll on 2015-09-18 23:56:26