In modular arithmetic, why is (x mod n)^y mod n == x^y mod n?
up vote
0
down vote
favorite
Why is (x mod n)^y mod n == x^y mod n? It seems to me like there is a property in modular arithmetic that explains why, does anyone have a simple way of explaining it?
modular-arithmetic
asked Nov 13 at 12:50
J. Doe
1
add a comment |
up vote
0
down vote
favorite
Why is (x mod n)^y mod n == x^y mod n? It seems to me like there is a property in modular arithmetic that explains why, does anyone have a simple way of explaining it?
modular-arithmetic
asked Nov 13 at 12:50
J. Doe
1
2
Hint: just show that $(x+n)^yequiv x^ypmod n$. The binomial theorem might help.
– lulu
Nov 13 at 12:53
It's possible that your comment made me understand it. Will let it bake a bit, and see if I connected the dots.
– J. Doe
Nov 13 at 13:56
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Why is (x mod n)^y mod n == x^y mod n? It seems to me like there is a property in modular arithmetic that explains why, does anyone have a simple way of explaining it?
modular-arithmetic
asked Nov 13 at 12:50
J. Doe
1
Why is (x mod n)^y mod n == x^y mod n? It seems to me like there is a property in modular arithmetic that explains why, does anyone have a simple way of explaining it?
modular-arithmetic
modular-arithmetic
asked Nov 13 at 12:50
J. Doe
1
asked Nov 13 at 12:50
J. Doe
1
asked Nov 13 at 12:50
J. Doe
1
asked Nov 13 at 12:50
J. Doe
1
asked Nov 13 at 12:50
J. Doe
1
1
2
Hint: just show that $(x+n)^yequiv x^ypmod n$. The binomial theorem might help.
– lulu
Nov 13 at 12:53
It's possible that your comment made me understand it. Will let it bake a bit, and see if I connected the dots.
– J. Doe
Nov 13 at 13:56
add a comment |
2
Hint: just show that $(x+n)^yequiv x^ypmod n$. The binomial theorem might help.
– lulu
Nov 13 at 12:53
It's possible that your comment made me understand it. Will let it bake a bit, and see if I connected the dots.
– J. Doe
Nov 13 at 13:56
2
2
Hint: just show that $(x+n)^yequiv x^ypmod n$. The binomial theorem might help.
– lulu
Nov 13 at 12:53
Hint: just show that $(x+n)^yequiv x^ypmod n$. The binomial theorem might help.
– lulu
Nov 13 at 12:53
It's possible that your comment made me understand it. Will let it bake a bit, and see if I connected the dots.
– J. Doe
Nov 13 at 13:56
It's possible that your comment made me understand it. Will let it bake a bit, and see if I connected the dots.
– J. Doe
Nov 13 at 13:56
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Let $$g^{k_1}=nq_1+r$$therefore$$(g^{k1}mod n)^{k2} mod n=r^{k_2}mod n$$Furthermore:$$(g^{k_1})^{k_2}mod n=(nq+r)^{k_2}mod n=nq_2+r^{k_2}mod n=r^{k_2}mod n$$therefore$$(g^{k1}mod n)^{k2} mod n=(g^{k_1})^{k_2}mod n$$
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $$g^{k_1}=nq_1+r$$therefore$$(g^{k1}mod n)^{k2} mod n=r^{k_2}mod n$$Furthermore:$$(g^{k_1})^{k_2}mod n=(nq+r)^{k_2}mod n=nq_2+r^{k_2}mod n=r^{k_2}mod n$$therefore$$(g^{k1}mod n)^{k2} mod n=(g^{k_1})^{k_2}mod n$$
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
add a comment |
up vote
0
down vote
Let $$g^{k_1}=nq_1+r$$therefore$$(g^{k1}mod n)^{k2} mod n=r^{k_2}mod n$$Furthermore:$$(g^{k_1})^{k_2}mod n=(nq+r)^{k_2}mod n=nq_2+r^{k_2}mod n=r^{k_2}mod n$$therefore$$(g^{k1}mod n)^{k2} mod n=(g^{k_1})^{k_2}mod n$$
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $$g^{k_1}=nq_1+r$$therefore$$(g^{k1}mod n)^{k2} mod n=r^{k_2}mod n$$Furthermore:$$(g^{k_1})^{k_2}mod n=(nq+r)^{k_2}mod n=nq_2+r^{k_2}mod n=r^{k_2}mod n$$therefore$$(g^{k1}mod n)^{k2} mod n=(g^{k_1})^{k_2}mod n$$
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
Let $$g^{k_1}=nq_1+r$$therefore$$(g^{k1}mod n)^{k2} mod n=r^{k_2}mod n$$Furthermore:$$(g^{k_1})^{k_2}mod n=(nq+r)^{k_2}mod n=nq_2+r^{k_2}mod n=r^{k_2}mod n$$therefore$$(g^{k1}mod n)^{k2} mod n=(g^{k_1})^{k_2}mod n$$
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
answered Nov 13 at 12:54
Mostafa Ayaz
11.7k3733
11.7k3733
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2996685%2fin-modular-arithmetic-why-is-x-mod-ny-mod-n-xy-mod-n%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
Hint: just show that $(x+n)^yequiv x^ypmod n$. The binomial theorem might help.
– lulu
Nov 13 at 12:53
It's possible that your comment made me understand it. Will let it bake a bit, and see if I connected the dots.
– J. Doe
Nov 13 at 13:56