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The number 1729 is famously the smallest positive integer expressible as the sum of two positive cubes in two different ways ($1729=1^3+12^3=9^3+10^3$). There is plenty of work on "taxicab numbers" - the smallest sums of cubes in $n$ different ways (which always exist) - Here's Ivars Peterson at MAA And here's another detailed analysis. (Does anyone know anything about the "Bill Butler" referred to in the second article)

However the sequence which caught my attention is OEIS A016078 - 4, 50, 1729, 635318657 which gives the smallest numbers which are sums of positive $n^{th}$ powers in two ways. Is there any more recent work or prospect of identifying such numbers for fifth powers and above? And should they be named as in the title of this post?

[This question arises from a much more frivolous one, which was closed, in which I learned why a $50^{th}$ birthday was special in this particular way].

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Just a note, there is no upper bound on number of ways to write as the sum of two squares, and those with exactly two ways as $x^2 + y^2, \; 1 \leq x \leq y$ are in a small number of infinite families. For cubes, fourth and fifth powers, it is not clear that any examples with three or more ways are known...wait, P. Vojta 1983 $$ 15170836545 = 517^3 + 2468^3 = 709^3 + 2456^3 = 1733^3 + 2152^3. $$ Euler $$635318657 = 133^4 + 134^4 = 59^4 + 158^4, $$ no triples known. There are parametric families that give some of these pairs. –  Will Jagy Oct 31 '12 at 21:45
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oeis.org/A046881 has better comments. –  SiliconCelery Oct 31 '12 at 21:46
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Alrighty, the thing about the Vojta example with cubes is that $\gcd(x,y) = 1$ for each pair. Hardy and Wright, Theorem 412 on page 333 in the Fifth edition paperback, show that you can get at least $r$ different ways for any $r$ if you do not restrict $\gcd(x,y). $ –  Will Jagy Oct 31 '12 at 21:54
    
@WillJagy Thanks for your comments on this and my previous. I have Hardy and Wright Fifth Edition, and will consult it. –  Mark Bennet Oct 31 '12 at 22:08
    
@SiliconCelery Thanks for that - very helpful. –  Mark Bennet Oct 31 '12 at 22:15

2 Answers 2

up vote 6 down vote accepted

Guy, Unsolved Problems In Number Theory, 3rd edition, D1, writes, "... it is not known if there is any nontrivial solution of $a^5+b^5=c^5+d^5$. Dick Lehmer once thought that there might be a solution with a sum of about 25 decimal digits, but a search by Blair Kelly yielded no nontrivial solution with sum $\le1.02\times10^{26}$."

At F30, Guy writes, "... $x^5$ is a likely answer to the following unsolved problem of Erdos. Find a polynomial $P(x)$ such that all the sums $P(a)+P(b)$ ($0\le a\lt b$) are distinct."

The book was published in 2004. I don't know whether there has been any progress since.

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Wikipedia is also not aware of any progress. –  joriki Oct 31 '12 at 22:09
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Guy's book is one I should have anyway, and it is missing from my collection. I have always been intrigued, because Hardy & Wright spend a fair amount of time on Waring's Problem (representation of integers as sums of $n^{th}$ powers) - but this seems to have been a bit of dead end - at least in the twentieth century - so far as deep mathematical insight and progress is involved. –  Mark Bennet Oct 31 '12 at 22:13
    
@joriki, true --- and Wikipedia's only citation is Guy's book. –  Gerry Myerson Nov 1 '12 at 0:12
    
Are $a$ and $b$ real and was there any recent progress? –  draks ... Nov 26 '13 at 23:10
    
@draks..., in the Erdos question at F30 in Guy, $a$ and $b$ are meant to be whole numbers. I'm not aware of any recent progress. –  Gerry Myerson Nov 27 '13 at 2:35

It is necessary to solve the equation:

$$x^5+y^5+z^5=q^5$$

For integers complex numbers solutions exist. $j=\sqrt{-1}$

Making this change.

$$a=p^2-2ps-s^2$$

$$b=p^2+2ps-s^2$$

$$c=p^2+s^2$$

You can write the solution.

$$x=jc+b$$

$$y=jc-b$$

$$z=a-jc$$

$$q=a+jc$$

$p,s$ - integers.

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