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Let a number n $ \in\mathbb Z,$ be expressible as the sum of its digits each raised to p $ \in\mathbb N$ . How to prove or disprove that there are infinite such numbers for a particular p?

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2 Answers 2

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For each $p$ there is a finite set of integers equal to the sum of its digits. Consider for $p=1,$ if $19\leq n\leq 99$ then $n$ has $2$ digits, whose sum can be at most $9+9=18 < n.$ If $100 \leq n \leq 999$ then $n$ has maximal digit sum $27 < n.$ It is easy to see this pattern is repeated and we never have an integer equal to its digit sum for $n\geq 19.$

If $p=2$ then for all $ 244 \leq n \leq 999$ the sum of the squares of the digits of $n$ is at most $9^2 + 9^2 +9^2 = 243 < n.$ For $1000\leq n \leq 9999$ then sum of the squares of the digits is at most $4 \cdot 9^2 = 324 < n.$ Clearly, for all $n> 243,$ $n$ is not the sum of the squares of its digits.

I leave it as an exercise to extend this idea for general $p.$

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If n is m digit, then the sum is at most $m9^p$ , so there is no n>$m9^p$, for which the situation is possible. OK? –  Swapnanil Saha Aug 17 '12 at 14:37

For more on this, see https://oeis.org/A023052 and references there.

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