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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.


Stage: 5 Challenge Level: Challenge Level:1

This is another solution by Yatir from Maccabim-Reut High School, Israel.

First let's prove that the sum of a positive number and its reciprocal is at least 2. It is clearly true for all x that $(x-1)^2 \geq 0$, hence $x^2-2x+1 \geq 0$ and it follows that $x^2 + 1 \geq 2x$. Since $x> 0$ there is no harm in dividing by $x$ proving that for all $x> 0$

$$x + {1\over x} \geq 2.$$

Another challenge is to use the hints given by the two illustrations in the question and to give alternative proofs that the sum of a positive number and its reciprocal is greater than or equal to 2.

Yatir uses this inequality when he sums k fractions and their reciprocals in the following proof. Can you use a similar method to give a shorter proof of the result without resorting to mathematical induction? You will need to expand the expression given in (1), collect pairs of terms, decide how many pairs there are and use the inequality for each of the pairs of terms.

We have to prove that:

$$(x_1 + x_2 + \cdots + x_n)({1\over x_1}+ {1\over x_2}+ \cdots + {1\over x_n}) \geq n^2.\quad (1)$$

Let's carry on by induction, first for $n=1$ we have $x_1 \times {1\over x_1} = 1 \geq 1^2$ so this part is proved.

Let's assume that it is true for a positive integer $n=k$: $$(x_1 + x_2 + \cdots + x_k)({1\over x_1}+ {1\over x_2}+ \cdots + {1\over x_k}) \geq k^2.$$

Now let's prove it for $n=k+1$. We have

$$\eqalign { (x_1 + x_2 + \cdots + x_k + x_{k+1})({1\over x_1}+ {1\over x_2}+ \cdots + {1\over x_k} + {1\over x_{k+1}}) &= (x_1 + x_2 + \cdots + x_k)({1\over x_1}+ {1\over x_2}+ \cdots + {1\over x_k})\cr &+ x_{k+1}({1\over x_1}+ {1\over x_2}+ \cdots + {1\over x_k})\cr &+ {1\over x_{k+1}}(x_1 + x_2 + \cdots + x_k)\cr &+ 1 \cr &\geq k^2 + \sum_{i=1}^k({x_{k+1}\over x_i}+ {x_i\over x_{k+1}})+ 1. \cr }$$

We have used the induction assumption that the result is true for $n=k$ in the last line. By opening this up we see that we have the sum of $k$ positive numbers and their reciprocals. Since each one of them is at least 2 we replace each of them by 2 so the inequality becomes:

$$\eqalign { (x_1 + x_2 + \cdots + x_k + x_{k+1})({1\over x_1}+ {1\over x_2}+ \cdots + {1\over x_k} + {1\over x_{k+1}}) \geq k^2 + 2k + 1 = (k+1)^2 \cr }$$

This proves the inequality for $n=k+1$ and so, by the axiom of induction, the result is true for all positive integers.