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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?

Exhaustion

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

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.

Inner Equality

Age 16 to 18 Short
Challenge Level
Well done to Amrit, Adithya, Daven and Sergio who all sent in solutions to this problem. Here are the first four inequalities:

$$ 10 < a+ b- c - d < 20 $$
$$ 0 < a- c < 10 $$
$$ -10 < a - c + d - b < 10 $$
$$ 0 < abcd < 625 $$
Aditha's solution explains how to get each of them, you can read the pdf

Amrit explained how to work out the fifth inequality:

For the last inequality, we need to prove the AM-GM inequality
$\frac{a+c}{2}>\sqrt{ac}$
$a+c>2\sqrt{ac}$
$a^2+2ac+c^2>4ac$
$a^2-2ac+c^2>0$
$a-c)^2>0$
A number squared is always greater than 0 unless the number is 0 or in this case if a=c
Plugging in a=c into the last inequality, we have
$\frac{|a|+|c|}{2}-\sqrt|ac|>0$
Looking at the AM-GM inequality, we want a and c to be as far apart as
possible. So |a| has to be 5 and |c| has to be 0 or vice versa.
Applying this, we have $\frac{|a|+|c|}{2}-\sqrt|ac|<\frac{5}{2}$