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Shades of Fermat's Last Theorem

The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?

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Exhaustion

Find the positive integer solutions of the equation (1+1/a)(1+1/b)(1+1/c) = 2

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Code to Zero

Find all 3 digit numbers such that by adding the first digit, the square of the second and the cube of the third you get the original number, for example 1 + 3^2 + 5^3 = 135.

Diverging

Stage: 5 Challenge Level: Challenge Level:1

Show that for natural numbers $x$ and $y$ if ${x\over y}> 1$ then

$${x\over y}> {(x+1)\over(y+1)}> 1.$$

Hence prove that

$$P = {2\over 1}{\cdot}{4\over 3}{\cdot}{6\over 5}{\cdots} {k\over k-1}> \sqrt{k+1}.$$

This shows that the product $P=\prod_{i=1}^n{2i\over{2i-1}}$ tends to infinity as $n$ tends to infinity. Now, using a similar method, show that

$$Q = {2\over 1}{\cdot}{4\over 3}{\cdot}{6\over 5}{\cdots}{100\over 99}> 12.$$