Does there exist $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$?
Does there exist $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$? Use the work from part a to justify the answer.
So in part a, I had to prove:
For each $[a] in mathbb Z_5 ,$ if $[a] neq [0] $, then $[a]^2 = [1]$ or $[a]^2=[4]$.
I was able to this, and determined that this was true for all of the cases.
Using the information I found from that part of the question though, I have to prove if there exists $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$.
All that I have been able to write down for my answer so far is:
For each integer $a$, if $ a ncong 0 (mod5)$, then $a^2 cong 1 (mod5)$ or $ a^2 cong 4 (mod5)$ and integer $n$ is a perfect square if $exists k in mathbb Z : n=k^2.$
I'm not sure where to go from here, any help would be appreciated!
equivalence-relations
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
asked Nov 30 '18 at 2:40
ClaireClaire
556
|
show 1 more comment
Does there exist $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$? Use the work from part a to justify the answer.
So in part a, I had to prove:
For each $[a] in mathbb Z_5 ,$ if $[a] neq [0] $, then $[a]^2 = [1]$ or $[a]^2=[4]$.
I was able to this, and determined that this was true for all of the cases.
Using the information I found from that part of the question though, I have to prove if there exists $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$.
All that I have been able to write down for my answer so far is:
For each integer $a$, if $ a ncong 0 (mod5)$, then $a^2 cong 1 (mod5)$ or $ a^2 cong 4 (mod5)$ and integer $n$ is a perfect square if $exists k in mathbb Z : n=k^2.$
I'm not sure where to go from here, any help would be appreciated!
equivalence-relations
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
asked Nov 30 '18 at 2:40
ClaireClaire
556
Well, if such an $a$ existed then $a^2$ is $3$ modulo $5$. Do you see the problem?
– T. Bongers
Nov 30 '18 at 2:46
where did you get the 3 from ? @T.Bongers
– Claire
Nov 30 '18 at 2:51
@Claire you number $5,158,232,468,953,153$ end with digit $3$.
– achille hui
Nov 30 '18 at 2:53
Is $5,158,232,468,953,153cong 1,4 pmod 5$?
– fleablood
Nov 30 '18 at 2:53
can any square number have 3 as the last digit.
– karakfa
Nov 30 '18 at 2:53
|
show 1 more comment
Does there exist $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$? Use the work from part a to justify the answer.
So in part a, I had to prove:
For each $[a] in mathbb Z_5 ,$ if $[a] neq [0] $, then $[a]^2 = [1]$ or $[a]^2=[4]$.
I was able to this, and determined that this was true for all of the cases.
Using the information I found from that part of the question though, I have to prove if there exists $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$.
All that I have been able to write down for my answer so far is:
For each integer $a$, if $ a ncong 0 (mod5)$, then $a^2 cong 1 (mod5)$ or $ a^2 cong 4 (mod5)$ and integer $n$ is a perfect square if $exists k in mathbb Z : n=k^2.$
I'm not sure where to go from here, any help would be appreciated!
equivalence-relations
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
asked Nov 30 '18 at 2:40
ClaireClaire
556
Does there exist $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$? Use the work from part a to justify the answer.
So in part a, I had to prove:
For each $[a] in mathbb Z_5 ,$ if $[a] neq [0] $, then $[a]^2 = [1]$ or $[a]^2=[4]$.
I was able to this, and determined that this was true for all of the cases.
Using the information I found from that part of the question though, I have to prove if there exists $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$.
All that I have been able to write down for my answer so far is:
For each integer $a$, if $ a ncong 0 (mod5)$, then $a^2 cong 1 (mod5)$ or $ a^2 cong 4 (mod5)$ and integer $n$ is a perfect square if $exists k in mathbb Z : n=k^2.$
I'm not sure where to go from here, any help would be appreciated!
equivalence-relations
equivalence-relations
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
asked Nov 30 '18 at 2:40
ClaireClaire
556
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
asked Nov 30 '18 at 2:40
ClaireClaire
556
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
edited Nov 30 '18 at 3:37
Robert Howard
1,9161822
1,9161822
asked Nov 30 '18 at 2:40
ClaireClaire
556
asked Nov 30 '18 at 2:40
ClaireClaire
556
asked Nov 30 '18 at 2:40
ClaireClaire
556
556
Well, if such an $a$ existed then $a^2$ is $3$ modulo $5$. Do you see the problem?
– T. Bongers
Nov 30 '18 at 2:46
where did you get the 3 from ? @T.Bongers
– Claire
Nov 30 '18 at 2:51
@Claire you number $5,158,232,468,953,153$ end with digit $3$.
– achille hui
Nov 30 '18 at 2:53
Is $5,158,232,468,953,153cong 1,4 pmod 5$?
– fleablood
Nov 30 '18 at 2:53
can any square number have 3 as the last digit.
– karakfa
Nov 30 '18 at 2:53
|
show 1 more comment
Well, if such an $a$ existed then $a^2$ is $3$ modulo $5$. Do you see the problem?
– T. Bongers
Nov 30 '18 at 2:46
where did you get the 3 from ? @T.Bongers
– Claire
Nov 30 '18 at 2:51
@Claire you number $5,158,232,468,953,153$ end with digit $3$.
– achille hui
Nov 30 '18 at 2:53
Is $5,158,232,468,953,153cong 1,4 pmod 5$?
– fleablood
Nov 30 '18 at 2:53
can any square number have 3 as the last digit.
– karakfa
Nov 30 '18 at 2:53
Well, if such an $a$ existed then $a^2$ is $3$ modulo $5$. Do you see the problem?
– T. Bongers
Nov 30 '18 at 2:46
Well, if such an $a$ existed then $a^2$ is $3$ modulo $5$. Do you see the problem?
– T. Bongers
Nov 30 '18 at 2:46
where did you get the 3 from ? @T.Bongers
– Claire
Nov 30 '18 at 2:51
where did you get the 3 from ? @T.Bongers
– Claire
Nov 30 '18 at 2:51
@Claire you number $5,158,232,468,953,153$ end with digit $3$.
– achille hui
Nov 30 '18 at 2:53
@Claire you number $5,158,232,468,953,153$ end with digit $3$.
– achille hui
Nov 30 '18 at 2:53
Is $5,158,232,468,953,153cong 1,4 pmod 5$?
– fleablood
Nov 30 '18 at 2:53
Is $5,158,232,468,953,153cong 1,4 pmod 5$?
– fleablood
Nov 30 '18 at 2:53
can any square number have 3 as the last digit.
– karakfa
Nov 30 '18 at 2:53
can any square number have 3 as the last digit.
– karakfa
Nov 30 '18 at 2:53
|
show 1 more comment
2 Answers
2
active
oldest
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So start by writing down the squares of the integers from $1,ldots,10$. Note that they end in only one of ${1,4,6,9}$. It is also easy to see that the last digit of a perfect square depends only on the last digit of its square root (write the square root as $10k+a$). You'll then see that any perfect square ends in one of the above digits, and 3 is not one of them. It also follows from the fact that you already found out that a perfect square is congruent module $1$ or $4$ to 5, which is essentially the same statement. Clearly, any integer which has $3$ as its last digit is congruent modulo $3$, which is a contradiction. PS: integers which are multiples of $10$ end with a $0$ (in fact, two of them), but I assume you aren't considering squares of $0$ when you look at $mathbb Z_5$.
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
add a comment |
You yourself wrote down
For each integer a,if a≆0 (mod 5),then a^2≅1(mod 5) or a^2≅4(mod 5)
Does $a^2 = 5,158,232,468,953,153$ satisfy those conditions?
However you have to consider that maybe $a cong 0 pmod 5$. In which case $a^2 cong 0 pmod 5$. Do $a^2$ satisfy that condition either?
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
add a comment |
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2 Answers
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2 Answers
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So start by writing down the squares of the integers from $1,ldots,10$. Note that they end in only one of ${1,4,6,9}$. It is also easy to see that the last digit of a perfect square depends only on the last digit of its square root (write the square root as $10k+a$). You'll then see that any perfect square ends in one of the above digits, and 3 is not one of them. It also follows from the fact that you already found out that a perfect square is congruent module $1$ or $4$ to 5, which is essentially the same statement. Clearly, any integer which has $3$ as its last digit is congruent modulo $3$, which is a contradiction. PS: integers which are multiples of $10$ end with a $0$ (in fact, two of them), but I assume you aren't considering squares of $0$ when you look at $mathbb Z_5$.
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
add a comment |
So start by writing down the squares of the integers from $1,ldots,10$. Note that they end in only one of ${1,4,6,9}$. It is also easy to see that the last digit of a perfect square depends only on the last digit of its square root (write the square root as $10k+a$). You'll then see that any perfect square ends in one of the above digits, and 3 is not one of them. It also follows from the fact that you already found out that a perfect square is congruent module $1$ or $4$ to 5, which is essentially the same statement. Clearly, any integer which has $3$ as its last digit is congruent modulo $3$, which is a contradiction. PS: integers which are multiples of $10$ end with a $0$ (in fact, two of them), but I assume you aren't considering squares of $0$ when you look at $mathbb Z_5$.
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
add a comment |
So start by writing down the squares of the integers from $1,ldots,10$. Note that they end in only one of ${1,4,6,9}$. It is also easy to see that the last digit of a perfect square depends only on the last digit of its square root (write the square root as $10k+a$). You'll then see that any perfect square ends in one of the above digits, and 3 is not one of them. It also follows from the fact that you already found out that a perfect square is congruent module $1$ or $4$ to 5, which is essentially the same statement. Clearly, any integer which has $3$ as its last digit is congruent modulo $3$, which is a contradiction. PS: integers which are multiples of $10$ end with a $0$ (in fact, two of them), but I assume you aren't considering squares of $0$ when you look at $mathbb Z_5$.
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
So start by writing down the squares of the integers from $1,ldots,10$. Note that they end in only one of ${1,4,6,9}$. It is also easy to see that the last digit of a perfect square depends only on the last digit of its square root (write the square root as $10k+a$). You'll then see that any perfect square ends in one of the above digits, and 3 is not one of them. It also follows from the fact that you already found out that a perfect square is congruent module $1$ or $4$ to 5, which is essentially the same statement. Clearly, any integer which has $3$ as its last digit is congruent modulo $3$, which is a contradiction. PS: integers which are multiples of $10$ end with a $0$ (in fact, two of them), but I assume you aren't considering squares of $0$ when you look at $mathbb Z_5$.
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
edited Nov 30 '18 at 5:39
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
answered Nov 30 '18 at 2:52
BoshuBoshu
705315
705315
add a comment |
add a comment |
You yourself wrote down
For each integer a,if a≆0 (mod 5),then a^2≅1(mod 5) or a^2≅4(mod 5)
Does $a^2 = 5,158,232,468,953,153$ satisfy those conditions?
However you have to consider that maybe $a cong 0 pmod 5$. In which case $a^2 cong 0 pmod 5$. Do $a^2$ satisfy that condition either?
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
add a comment |
You yourself wrote down
For each integer a,if a≆0 (mod 5),then a^2≅1(mod 5) or a^2≅4(mod 5)
Does $a^2 = 5,158,232,468,953,153$ satisfy those conditions?
However you have to consider that maybe $a cong 0 pmod 5$. In which case $a^2 cong 0 pmod 5$. Do $a^2$ satisfy that condition either?
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
add a comment |
You yourself wrote down
For each integer a,if a≆0 (mod 5),then a^2≅1(mod 5) or a^2≅4(mod 5)
Does $a^2 = 5,158,232,468,953,153$ satisfy those conditions?
However you have to consider that maybe $a cong 0 pmod 5$. In which case $a^2 cong 0 pmod 5$. Do $a^2$ satisfy that condition either?
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
You yourself wrote down
For each integer a,if a≆0 (mod 5),then a^2≅1(mod 5) or a^2≅4(mod 5)
Does $a^2 = 5,158,232,468,953,153$ satisfy those conditions?
However you have to consider that maybe $a cong 0 pmod 5$. In which case $a^2 cong 0 pmod 5$. Do $a^2$ satisfy that condition either?
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
answered Nov 30 '18 at 2:57
fleabloodfleablood
68.5k22685
68.5k22685
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
add a comment |
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
No, because either way, the number doesn't end in a 0 or a 5, so it's not divisible by 5. The 0 condition isn't satisified either because perfect squares only end in 1, 4, 6, or 9
– Claire
Nov 30 '18 at 2:59
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Is this the reason why it's not true then? @fleablood
– Claire
Nov 30 '18 at 3:18
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
Well, you DID say $a^2$ must be equiv $1$ or $4$ mod $5$. Or $a$ equiv $0$. And $5,1....,.53$ is not that. So you tell me. Is it possible for something to happen that you yourself proved could not happen?
– fleablood
Nov 30 '18 at 4:27
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
But perfect squares like 100, 400, 900, 1600, etc. can and with $0$.
– fleablood
Nov 30 '18 at 4:28
add a comment |
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Well, if such an $a$ existed then $a^2$ is $3$ modulo $5$. Do you see the problem?
– T. Bongers
Nov 30 '18 at 2:46
where did you get the 3 from ? @T.Bongers
– Claire
Nov 30 '18 at 2:51
@Claire you number $5,158,232,468,953,153$ end with digit $3$.
– achille hui
Nov 30 '18 at 2:53
Is $5,158,232,468,953,153cong 1,4 pmod 5$?
– fleablood
Nov 30 '18 at 2:53
can any square number have 3 as the last digit.
– karakfa
Nov 30 '18 at 2:53