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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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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
Well done to Minhaj, Amrit, Adithya, Sasi, Will, Charlie, Luke and Luke, Eric, Pablo, Julian, Johnny, Peter, Gabriel, and Manolis for their hard work on this problem!

Peter made the following observation:

I worked out that the range of numbers is one to three if you are talking about positive intergers.

Gabriel used a graphical method to solve the inequality.

Others used variations of the following method:

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 $0.072< x< 13.928.$

Click on the pdf links to read Eric's, Minhaj's, Manolis's and Sasi's solutions.