ok, so problem solved right

Regarding sufficient condition, request explanation.

do you mean "if for $x$, a^x \equiv a^r \pmod{n}$ then $a^n \equiv 1 \pmod{n}$?

How is that same as the sufficient condition?

$a^x\%n=(a^n\%n)^{x/n}(a^{x\%n}\%n)\%n= 1.a^{x\%n}$.

$a^x\%n=(a^n\%n)^{q}(a^{r}\%n)\%n$

How to equate lhs and rhs, and how is above a sufficient condition?

try not to say "just sufficient condition", sufficient condition of what?

i can't see the big picture, what are you trying to do?

Sufficient condition for given to be true.

I.e. $x^n \equiv 1 \pmod n$

In fact am unclear about what's happening.

Ian wrote before, but very /totally unclear by what he means : What I was saying before was that if an≢1modn then "probably" amn≢1modn also. If this happens then I expected that there would be no way to have ax≡ax%nmodn, but actually there are some other possibilities, such as ax%n≡0modn. (This is the only other possibility if n is prime, but there might be another way if n is composite.)

perhaps you would like to write down what you are trying to prove first

$a^n \equiv 1 \pmod{n}$ if and only if for all $x, a^x \equiv a^r \pmod{n}$

this is what I understand

Seems are stating the necessary part as stated by Ian, not the sufficient part as stated above.

I stated an if and only if statement

I understand you mean:

$a^{x-r} \pmod{n} \equiv 1 \pmod{n}$ $ \iff a^{qn} \equiv 1 \pmod{n}$

But is that the same as the sufficient condition.

You can help too much easily. Still feel lost in this question??

sufficient condition for ?

perhaps u can engage him?

i do not know what you are trying to do

Not possible, engaged him many times. You can see it.

If can understand from what he stated, then will get all.

what are the statements that you are trying to find the sufficient condition to?

as of now if you can't answer that question, you are asking me what is the answer without telling me what is the question

$x^n \equiv 1 \pmod n$

Sufficient condition for : $x^n \equiv 1 \pmod n$

sufficient condition to what?

$x^n \equiv 1 \pmod n$

huh

If am wrong, then that is the reason stopped asking him. No benefit as nothing added to understanding.

Ian wrote before, but very /totally unclear by what he means : What I was saying before was that if $a^n≢1\pmod n$ then "probably" $a^{mn}≢1\pmod n$ also.// Don't know what he means by "probably".

means it is not always true

That he wrote in context of his earlier (and first )comment:

that is why i process it and include a quantifier statement

That he wrote in context of his first comment: The point is that if$ a^n≢1\pmod n$ then the $mn$ part in the $x$ can change the remainder.

Please tell me what are the necessary and sufficient condition for

I mean your iff condition can understand, and $\forall x$ quantifier too, but what is it for Ian states necessary and sufficient conditions?

i suspect what i stated is what he claimed

Please elaborate.

$a^n \equiv 1 \pmod{n}$ if and only if for all $x,a^x≡a^r\pmod{n}$

I think he consider $a^n \equiv 1 \pmod{n}$ and $a^n \not \equiv 1 \pmod{n}$, two cases right?

hence he covers both cases

since he considers two cases, he has found the sufficient and necessary condition for all $x$, $a^x \equiv a^r \pmod{n}$.

Where is this consideration of both cases (not like your answer, where you consider only one case of $a^n \equiv 1 \pmod{n}$ ) specified in his (Ian) statement of sufficient and necessary conditions?

he considers $a^n \equiv 1$ and $a^n \not \equiv 1$ separately right

"if $a^n \equiv 1$, that would be the sufficient condition for $a^x \equiv a^r \pmod{n}$.

and then after which he check $a^n \not \equiv 1$ to check that $a^n \equiv 1$ is also a necessary condition

that is it is an equivalence statement

Please elaborate with an example, as (what have construed) sufficient condition for success case is (sorry, if got wrong) necessary condition for failure case.

Please respond.

Without example, very difficult to understand.

If $a^n \equiv 1$, we proved that $a^x \equiv a^r \pmod{n}$. Hence $a^n \equiv 1$ is a sufficient condition for $a^x \equiv a^r \pmod{n}$.

I think need take two set of values for $a,x,n$ s.t. failure occurs in one case , I.e.$a^x\pmod n\equiv a^{x \%n} \pmod n$. While in other case, it fails. Then, need show that the sufficient condition for success case is necessary condition for the failure case.

Now if $a^n \not\equiv 1$, we proved that $a^x \equiv a^r \pmod{n}$ is not true for all $x$.

hence the only way for all $x$, such that $a^x \equiv a^r \pmod{n}$ to hold true is $a^n \equiv 1$.

hence $a^n \equiv 1$ is the sufficient and necessary condition for $a^x \equiv a^r \pmod{n}$.

Sorry, for going back to complete my last comment.

Take help of such set of values of $a,x,n$ to explain your point, please.

i am claiming a general statement

...

$a^n \equiv 1 \pmod{n} \iff \forall x, a^x \equiv a^r \pmod{n}$.

are you able to undersrtand the equivalence?

you can try with particular numbers to check if both take the same truth value

but i m having a universal statement on one side, so to verify it, you have to check all x

Yes, it was stated earlier too. You showed a new result on basis of that - for $a^n \equiv 1$ being the sufficient and necessary condition for $a^x \equiv a^r \pmod{n}$. But, there must be some difference between necessary and sufficient conditions!

i didn't think through carefully but i believed he gave a proof, please check his argument

did he claim any condition?

such as $n$ is prime?

oh well, it is beyond me.

I am not sure if what he claims is true for now

or if i have misunderstood him

Why prime value of $n$ is tough? You in earlier chat, referred to fermat's little theorem, for prime values. Though at that time, no details were given by you.

If $a^n \equiv 1 \pmod{n} \iff \forall x, a^x \equiv a^r \pmod{n}$. Take set of values: $n=3, x=2, a=2$. Here, $(a,n)=1$. Get: $2^3\equiv 2\pmod n \not\equiv 1 \pmod{n}$. And, $q=0, r=2$. I am trying natural values for $x$ just taking care of $(a,n)=1$.