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
    2






    active

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    active

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    active

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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

            68.7k22685




            68.7k22685












            • 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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            Ida-Boy-Ed-Garten

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            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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