Divisibility of fourths by seven The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraA question on primes and divisibilityDivisibility by 37 proofBasic divisibility of large numbers.How to prove divisibility of the difference between two numbers.Better Divisibility by 8How do you prove this divisibility?Proof that $2^222-1$ is divisible by 3Induction proof, divisibilityDivisibility and GCD proofTransitivity of the divisibility relationship among integers.
Match Roman Numerals
What force causes entropy to increase?
ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?
How to support a colleague who finds meetings extremely tiring?
Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?
What can I do if neighbor is blocking my solar panels intentionally?
What is the padding with red substance inside of steak packaging?
Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)
How do spell lists change if the party levels up without taking a long rest?
Are spiders unable to hurt humans, especially very small spiders?
A phrase ”follow into" in a context
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
What other Star Trek series did the main TNG cast show up in?
Can a flute soloist sit?
The following signatures were invalid: EXPKEYSIG 1397BC53640DB551
How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave?
How do you keep chess fun when your opponent constantly beats you?
Is every episode of "Where are my Pants?" identical?
What do I do when my TA workload is more than expected?
Does Parliament hold absolute power in the UK?
Intergalactic human space ship encounters another ship, character gets shunted off beyond known universe, reality starts collapsing
Is there a writing software that you can sort scenes like slides in PowerPoint?
Button changing its text & action. Good or terrible?
For what reasons would an animal species NOT cross a *horizontal* land bridge?
Divisibility of fourths by seven
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraA question on primes and divisibilityDivisibility by 37 proofBasic divisibility of large numbers.How to prove divisibility of the difference between two numbers.Better Divisibility by 8How do you prove this divisibility?Proof that $2^222-1$ is divisible by 3Induction proof, divisibilityDivisibility and GCD proofTransitivity of the divisibility relationship among integers.
$begingroup$
I am otherwise very good in mathematics, but recently I came upon a problem that I just can't solve. Do you have any idea how to solve it?
If $a^2 + b^2 + c^2$ is divisible by 7, prove that $a^4 + b^4 + c^4$ is divisible by 7 as well.
Thanks!
divisibility
asked Mar 31 at 7:15
PygmalionPygmalion
1286
$endgroup$
add a comment |
$begingroup$
I am otherwise very good in mathematics, but recently I came upon a problem that I just can't solve. Do you have any idea how to solve it?
If $a^2 + b^2 + c^2$ is divisible by 7, prove that $a^4 + b^4 + c^4$ is divisible by 7 as well.
Thanks!
divisibility
asked Mar 31 at 7:15
PygmalionPygmalion
1286
$endgroup$
add a comment |
$begingroup$
I am otherwise very good in mathematics, but recently I came upon a problem that I just can't solve. Do you have any idea how to solve it?
If $a^2 + b^2 + c^2$ is divisible by 7, prove that $a^4 + b^4 + c^4$ is divisible by 7 as well.
Thanks!
divisibility
asked Mar 31 at 7:15
PygmalionPygmalion
1286
$endgroup$
I am otherwise very good in mathematics, but recently I came upon a problem that I just can't solve. Do you have any idea how to solve it?
If $a^2 + b^2 + c^2$ is divisible by 7, prove that $a^4 + b^4 + c^4$ is divisible by 7 as well.
Thanks!
divisibility
divisibility
asked Mar 31 at 7:15
PygmalionPygmalion
1286
asked Mar 31 at 7:15
PygmalionPygmalion
1286
asked Mar 31 at 7:15
PygmalionPygmalion
1286
asked Mar 31 at 7:15
PygmalionPygmalion
1286
asked Mar 31 at 7:15
PygmalionPygmalion
1286
1286
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Modulo $7$, squares are congruent to $0$, $1$, $2$ or $4$. The only way a trio
of these can add to zero modulo $7$ is for all of them to be $0$, or for them
to be some permutation of $1$, $2$ and $4$. In the latter case, $a^4+b^4+c^4
equiv 1^2+2^2+4^2pmod 7$.
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
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%2f3169132%2fdivisibility-of-fourths-by-seven%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Modulo $7$, squares are congruent to $0$, $1$, $2$ or $4$. The only way a trio
of these can add to zero modulo $7$ is for all of them to be $0$, or for them
to be some permutation of $1$, $2$ and $4$. In the latter case, $a^4+b^4+c^4
equiv 1^2+2^2+4^2pmod 7$.
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
add a comment |
$begingroup$
Modulo $7$, squares are congruent to $0$, $1$, $2$ or $4$. The only way a trio
of these can add to zero modulo $7$ is for all of them to be $0$, or for them
to be some permutation of $1$, $2$ and $4$. In the latter case, $a^4+b^4+c^4
equiv 1^2+2^2+4^2pmod 7$.
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
add a comment |
$begingroup$
Modulo $7$, squares are congruent to $0$, $1$, $2$ or $4$. The only way a trio
of these can add to zero modulo $7$ is for all of them to be $0$, or for them
to be some permutation of $1$, $2$ and $4$. In the latter case, $a^4+b^4+c^4
equiv 1^2+2^2+4^2pmod 7$.
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
Modulo $7$, squares are congruent to $0$, $1$, $2$ or $4$. The only way a trio
of these can add to zero modulo $7$ is for all of them to be $0$, or for them
to be some permutation of $1$, $2$ and $4$. In the latter case, $a^4+b^4+c^4
equiv 1^2+2^2+4^2pmod 7$.
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 31 at 7:27
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
add a comment |
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
Is that a special, well-known rule that modulo 7 of squares is 0, 1, 2 and 4?
$endgroup$
– Pygmalion
Mar 31 at 7:38
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion I'd say it's well-known, but nothing special.
$endgroup$
– Lord Shark the Unknown
Mar 31 at 7:48
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@Pygmalion In a problem like this, it should be obvious that it is useful to know and it only takes a couple of minutes to check. I didn't happen to know that but I hope that I would have found it quickly.
$endgroup$
– badjohn
Mar 31 at 8:04
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@badjohn I suspected that there should be something special about divisibility of 7, but I only found another rule that claims that 10*a+b is divisible by 7 if a-2*b is divisible by 7. Obviously not very useful in this case.
$endgroup$
– Pygmalion
Mar 31 at 8:17
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
$begingroup$
@Pygmalion In a problem like this, with a small modulus, I would start by just writing out a table of squares.
$endgroup$
– badjohn
Mar 31 at 8:19
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3169132%2fdivisibility-of-fourths-by-seven%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
