Prove that any positive integer can be expressed as a combination of distinct powers of $\pm4^n$ and $\pm3^{2m+1}$(where m,n are integers). Like $33=4^3-4^0-3^3-3^1$
Prove any positive integer can be expressed as a combination of distinct powers of $\pm4^n$ and $\pm3^{2m-1}$
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number-theory
discrete-mathematics
integers
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0This is not so. False statement. – 2017-01-24
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2What do you mean by "combination"? – 2017-01-24
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0How do you get $5$? You don't allow $3^2-4^1$ or $4^1+3^0$ – 2017-01-24
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0just $4^0$+$4^1$, there's no restriction on n @RossMillikan – 2017-01-24
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0like add or subtract those powers @BillDubuque – 2017-01-24
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0Can you give a counterexample plz?@individ – 2017-01-24
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0It is not a question but an absurdity. Put any number of terms and having $a^0$ - you can write anything. In the original formulation dealt with the case. $p=\pm4^{n}\pm3^{2m+1}$ – 2017-01-24
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1But the restriction is to use distinct powers so that $a^0$ can only be used once, it cannot be proved by induction. @individ – 2017-01-24
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0I don't believe this can be easily proven (or disproven, for that matter). – 2017-01-24
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0A cool question. Where did you find it? Is this something you toyed with yourself, or is it from contest training or...? Give such context to prevent it from attracting too much negative attention, please. Or, if you have worked on it, please give details of your findings (integers you have trouble presenting in this form). – 2017-01-24
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0I might try to describe the base-4 "digits" of the set $S$ of positive integers that can be written using $\pm 4^n$. And then hope to show that the integers missing from that list can be "fixed" by adding/subtracting an allowed combination of powers of three. For example, if $n\in S$ then any string of 2s in its base-4 presentation must be followed by a 3, a 1 can only be followed by 0 or 1,... possibly other rules that I didn't see right away :-/. I share Ivan Neretin's view that this may be rather difficult. – 2017-01-24
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0Thanks for the inspiration. I toyed it myself and it's a problem I derive while working on Erdos number problems. I'm still working on it but no significant discovery right now.@JyrkiLahtonen – 2017-01-24