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?

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.

Challenge Level

Prove that in every tetrahedron there is a vertex such that the three edges meeting there have lengths which could be the sides of a triangle.

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NRICH is part of the family of activities in the Millennium Mathematics Project.

NRICH is part of the family of activities in the Millennium Mathematics Project.