## solving x^{n}+a=mod(b)

If a,b,n are integers..how could you solve:

$$x^{n}+a=mod(b)$$ n>0 and integer..

If possible please put an "step by step" example..i have managed to solve the linear congruence

$$ax=bmod(c)$$ but i don't know how to solve "upper" congruences..thanks.
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 Recognitions: Homework Help Science Advisor I really don't know what you're asking. I presume the question you've already solved is $$ax\equiv b\pmod c$$ but I'm not sure what your first is. Depending on what you mean, you might be able to use reciprocity.
 i know how to solve for n=1 then i would like to know how to solve the special case y proposed.... - By the way...are not there "approximate" or numerical method to calculate the set of the solutions?..

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## solving x^{n}+a=mod(b)

If you can solve for any x other than 1 (or 2) easily then you've just proven RSA easy to break. Come on, Jose, think for a second before posting a question: taking roots mod c is hard, and that's the whole point of RSA.
 - I don't know why is the "congruence" $$x^{n}=amod(b)$$ for n>1 in fact it's nothing but a Monomial..you could calculate all the roots of $$x^{n}-a=0$$ easily and from this find a solution...
 Recognitions: Homework Help Science Advisor Then why don't you 'just do it'? You really ought to get straight in your mind what a function is, what an algorithm to find a value of said function is, and what you consider to be a reasonable cost for carrying out the algorithm, and then state what that is.

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 -For example "Hallfosftivy"..a solution to the (general ) congruence: $$f(x)=amod(x)$$ (for example) is the solution to the "equation" (non-linear) in the form: $$[\frac{f(x)-a}{x}]-\frac{f(x)-a}{x}=0$$ or if we define the function: $$=x-[x]$$ where "[x] " is the floor function then: $$<\frac{f(x)-a}{x}>=0$$ so from this you could calculate all the "roots" (integers) and solution to the initial congruence...