What are the solutions in integers of $a^3+2a+1=2^b$?
[Source: Serbian competition problem]
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1Related link: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=152991& – JimmyK4542 Oct 28 '14 at 08:19
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We can discount any negative solutions, the LHS will be a negative integer and the RHS will be a positive rational. – Bennett Gardiner Oct 28 '14 at 08:22
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Other than $b=2$, there is no other with $2<b<1000$. – Tito Piezas III Nov 23 '14 at 00:20
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Does this answer your question? Equation $x^3+2x+1=2^n$ in positive integers – Apass.Jack Nov 22 '22 at 19:34
1 Answers
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My suggestion would be to start with a(a+2)=2b-1
We also notice that when a is odd, a2+2 is odd. likewise if a is even, a2+2 is even
i.) From here, we can generalize that since 2b-1 is odd for all integers a and b > 0, then a must always be oddii.) We notice that the last digit of 2b-1 ossiclates between the numbers 1,3,7,5
iii.)From ii.) the remaining values of the last digit of a can be reduced to 3, 5 and 7
(i.e. b mod 4= 0 or 2 ; b is a multiple of 2)
Combining all the conditions, we can form the diophantine equation: a(a2+2)=(2b+1)(2b-1)
drewyu
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As far as I can see, $2^b-1$ and $2^b+1$ are always consecutive odd numbers. – egreg Jun 25 '15 at 17:41
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You can say it in a more simple way: $a^3 + 2a \equiv 0\pmod 3$, $2^b\equiv (-1)^b\pmod 3$; therefore $1\equiv (-1)^b \pmod 3$, and $b=2b_1$. – Michael Galuza Jun 28 '15 at 08:23