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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.

In Between

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

There were many correct solutions to this problem: Alex from Stoke on Trent Sixth Form; Manuele and Eduardo, both from the British School, Manila; Jeff from Sierra Vista; Dapeng from Claremont Fancourt School, Diarmuid from Trinity School, Nottingham; Christina from Smithdon High School; Thomas from A Y Jackson School; Tathagata from South Point High School; Jack from Guthlaxton College; Innyeong Chang from Nanjing International School; Chen Jinsheng from ACS Independent School and John from Takapuna Grammar School.

This is Alex's solution:

Multiply each side of the inequality by $\sqrt x$. There is no problem about reversing the inequality sign, because $\sqrt x$ is given as positive. This gives: $x + 1 < 4\sqrt x$.
Subtract $4\sqrt x$ from both sides so the right-hand side is zero to give the following inequality involving a quadratic in $\sqrt x$:
$$x - 4\sqrt x + 1 < 0.$$
Use the quadratic formula to factorise the left-hand side:
$$(\sqrt x - 2 -\sqrt 3)(\sqrt x - 2 + \sqrt 3) < 0.$$
Considering the value of the left-hand side for the positive values of $\sqrt x$, the left-hand side is less than zero as required only for the range
$$2 - \sqrt 3 < \sqrt x < 2 + \sqrt 3.$$
Here $x$ and $\sqrt x$ are both positive, and each value of $\sqrt x$ corresponds to a sole $x$ value and vice-versa, so everything in the inequality can be squared without losing solutions. The solution of the inequality is:
$$7-4\sqrt 3 < x < 7+4\sqrt 3.$$
Note that $7-4\sqrt 3\approx 0.0718 > 0$.

Christina's method was slightly more direct:

Square both sides. Multiply throughout by x. Rearrange to form the quadratic inequality:
$$x^2-14x+1< 0.$$
Use the quadratic formula to solve this inequality. From the graph of $y =x^2 -14x + 1$ we see that the solution is $7-4\sqrt 3 < x < 7+4\sqrt 3$ or approximately $0.072< x< 13.928.$