Does there exist $a in mathbb Z$ such that $a^2 = 5,158, 232,468,953,153$?












1















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







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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
















1















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







share|cite|improve this question
























  • 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














1












1








1








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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 3:37









Robert Howard

1,9161822




1,9161822










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


















  • 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










2 Answers
2






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oldest

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1














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 $mathbb Z_5$.






share|cite|improve this answer































    0














    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?






    share|cite|improve this answer





















    • 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











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    2 Answers
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    2 Answers
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    1














    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 $mathbb Z_5$.






    share|cite|improve this answer




























      1














      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 $mathbb Z_5$.






      share|cite|improve this answer


























        1












        1








        1






        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 $mathbb Z_5$.






        share|cite|improve this answer














        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 $mathbb Z_5$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 5:39

























        answered Nov 30 '18 at 2:52









        BoshuBoshu

        705315




        705315























            0














            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?






            share|cite|improve this answer





















            • 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
















            0














            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?






            share|cite|improve this answer





















            • 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














            0












            0








            0






            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?






            share|cite|improve this answer












            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?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • 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


















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