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Find a natural number that cannot be written as the sum of at most 36 fifth powers.

This is a question from last years test on elementary number theory, but I have no clue on how to handle this. Any tips would be appreciated.

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    $\begingroup$ Just a guess, but maybe $1+\prod_{n=1}^{36} n^5$? $\endgroup$ – Mike Pierce Oct 24 '16 at 18:59
  • $\begingroup$ It's probably shown in: Chen, J.-R., Waring's Problem for g(5)=37. Sci. Sinica 13, 1547-1568, 1964. Also appeared as Chinese Math Acta 6, 105-127, 1965. (I found this reference at Wolfram MathWorld.) $\endgroup$ – Mitchell Spector Oct 24 '16 at 19:03
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$$ 223 $$

For Waring's problem, it is expected that $$ n = 2^k \left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor - 1 $$ achieves $g(k),$ as it can only be written using $1$ and $2^k,$ as it is smaller than $3^k.$ It is also one smaller than an integer multiple of $2^k,$ so the best that can be done is one fewer copies of $2^k,$ plus $2^k - 1$ copies of $1.$

Note $$ \left( \frac{3}{2} \right)^5 = 7.59375, $$ so $$ \left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor = 7. $$

This $n$ requires $$ 2^k + \left\lfloor \left( \frac{3}{2} \right)^k \right\rfloor - 2 $$ $k$th powers. So, $g(k) $ or not, this does give a lower bound for $g(k).$ Here, $$ 32 + 7 - 2 = 37. $$ $$ 223 = 6 \cdot 32 + 31 = 6 \cdot 2^5 + 31, $$ therefore needing $37$ fifth powers, as expected.

It is far more difficult to prove the value of $g(k),$ here $g(5).$ The problem does not require a proof, just one example that gives a lower bound, and probably shows the value of $g(k).$

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