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Is there a way of deducing the smallest integer values for $a, b$ and $c$ that satisfy either

$( a ^ b ) + c = d$

or

$( a ^ b ) - c = d$

such that the addition $( a + b + c )$ is the smallest possible integer?

I am wanting to do this for some very large integer values up to approximately 14 000 000 000 000 digits long. I may have to work with some smaller numbers to start with however.

I could write a computer program to determine some numbers but I was wondering if there is some higher level mathematics involved that could provide a solution please.

Vladhagen
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John Anthony Oliver
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  • For any positive $d$, $x^1-(x-d)=d$, so the required sum can be $x+1+(x-d)$. Now the sum can be arbitrarily negative by reducing $x$. – peterwhy Jan 23 '14 at 16:53
  • vadim123. Imagine if the number for d is now = 13579246812435687903215137539757 or any long number you care to imagine. What then are the smallest numbers for a, b and c respectively?

    I am looking for a way to see if they can be determined mathematically without using a computer program to do so that uses loops to find any solution.

     – John Anthony Oliver Jan 23 '14 at 16:55
  • peterwhy. What if there are no solutions for some values of d where c = 0 or even c = 1 ? – John Anthony Oliver Jan 23 '14 at 17:03
  • There is always a solution $(d-1)^1+1 = d$, so the sum is $(d-1)+1+1 = d+1$. – peterwhy Jan 23 '14 at 17:08

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If $b>1$ and $a^b\approx d$ then $(a\pm1)^b$ differs from $a$ by $\approx ba^{b-1}\approx\frac{bd}{a}$, hence we can achieve $c\approx \frac{bd}a$ or better. For given $b$, we will have $a\approx \sqrt[b]d$, hence $$ a+b+c\approx \sqrt[b]d +b+\frac{bd}{\sqrt[b]d}.$$ In this light, the best choce seems to be $b=2$ as it keeps the last summand small. Anything better than that is "luck", e.g. if $d$ happens to be awfully close to some high (odd) power. Computationally, it is not too problematic to compute $a=\lfloor \sqrt[b]d\rfloor $ for $b=2,3,\ldots$ and determine the $c$ belonging to $a^b$ and $(a+1)^b$ until you reach $a<10$ say. After that try the obvious best choices of $b,c$ for $a=2,\ldots , 9$ accordingly

Hagen von Eitzen
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  • Thank you for your reply. :-) I may have to test the idea programatically.

    However consider d = 13579246812435687903215137539757 ( that I mentioned in an earlier comment ).

    If b = 3685002959623735 then 3685002959623735 X 3685002959623735 = 13579246812435686322652775350225

    c = 13579246812435687903215137539757 - 13579246812435686322652775350225 = 1580562362189532

    You see that the value for c is still quite large. I will have to test other values for b to see how small c can become. [continued in next comment]

     – John Anthony Oliver Jan 23 '14 at 18:08
  • I may end up determining that the ideal value for b may be say 2 for a certain range of numbers and gradually higher as the value for d grows larger. – John Anthony Oliver Jan 23 '14 at 18:08