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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.

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

Graeme showed the inequalities for us:

$7 = 9^{1/2} + 8^{1/3} + 16^{1/4} > 7^{1/2} + 7^{1/3} + 7^{1/4}$

$4 = 4^{1/2} + 1^{1/3} + 1^{1/4} < 4^{1/2} + 4^{1/3} + 4^{1/4}$

While I was at it, I came up with these:

$6 = 6.25^{1/2} + 6.859^{1/3} + 6.5536^{1/4} > 6^{1/2} + 6^{1/3} + 6^{1/4}$

Although that looks hard, it can be done without a calculator by partitioning $6$ into $2.5+1.9+1.6$, and finding appropriate powers of each number. The last one is easy for computer geeks like me who have memorized many small powers of $2$.

$5 < 4^{1/2} + 4.096^{1/3} + 4^{1/4} < 5^{1/2} + 5^{1/3} + 5^{1/4}$

This, too, is pretty easy without a calculator - $4.096^{1/3}$ is $1.6$, and the square root of $2$ is more than $1.4$, so the first sum is more than $5$, and clearly less than the second sum.

Thanks for the extensions, Graeme.

These inequalities show that the graph is going to intersect the x-axis somewhere between 4 and 7, which it does: