Cubes and squares

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Consider there are five consecutive numbers $a,b,c,d$ and $e$ such that $a lt b lt c lt dlt e$.Given that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, find the least value of $c$.



I just want to know how to approach these kind of problems.



Thanx for any help. :)




square-numbers




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    So $b+c+d=3c$ and $a+b+c+d+e=5c$?
    – Lord Shark the Unknown
    Jul 29 at 7:33














up vote
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Consider there are five consecutive numbers $a,b,c,d$ and $e$ such that $a lt b lt c lt dlt e$.Given that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, find the least value of $c$.



I just want to know how to approach these kind of problems.



Thanx for any help. :)




square-numbers




share|cite|improve this question















  • 2




    So $b+c+d=3c$ and $a+b+c+d+e=5c$?
    – Lord Shark the Unknown
    Jul 29 at 7:33












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Consider there are five consecutive numbers $a,b,c,d$ and $e$ such that $a lt b lt c lt dlt e$.Given that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, find the least value of $c$.



I just want to know how to approach these kind of problems.



Thanx for any help. :)




square-numbers




share|cite|improve this question











Consider there are five consecutive numbers $a,b,c,d$ and $e$ such that $a lt b lt c lt dlt e$.Given that $b+c+d$ is a perfect square and $a+b+c+d+e$ is a perfect cube, find the least value of $c$.



I just want to know how to approach these kind of problems.



Thanx for any help. :)





square-numbers





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 29 at 7:26









Shukraditya Bose

215




215







  • 2




    So $b+c+d=3c$ and $a+b+c+d+e=5c$?
    – Lord Shark the Unknown
    Jul 29 at 7:33












  • 2




    So $b+c+d=3c$ and $a+b+c+d+e=5c$?
    – Lord Shark the Unknown
    Jul 29 at 7:33







2




2




So $b+c+d=3c$ and $a+b+c+d+e=5c$?
– Lord Shark the Unknown
Jul 29 at 7:33




So $b+c+d=3c$ and $a+b+c+d+e=5c$?
– Lord Shark the Unknown
Jul 29 at 7:33










1 Answer
1






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2
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Restating @Lord Shark's hint, we have that $3c$ must be a perfect square and $5c$ a perfect cube.



The smallest such number $c$ would obviously have prime factors of the form $3^x5^y$.



Now, $3^x+15^y$ is a perfect square $implies$ $x + 1$ and $y$ are even.



And, $3^x5^y+1$ is a perfect cube $implies$ $x$ and $y + 1$ are divisible by $3$.



Can you solve for the smallest such $x$ and $y$?






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  • thanx :) for the help
    – Shukraditya Bose
    Jul 29 at 13:20










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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes








up vote
2
down vote













Restating @Lord Shark's hint, we have that $3c$ must be a perfect square and $5c$ a perfect cube.



The smallest such number $c$ would obviously have prime factors of the form $3^x5^y$.



Now, $3^x+15^y$ is a perfect square $implies$ $x + 1$ and $y$ are even.



And, $3^x5^y+1$ is a perfect cube $implies$ $x$ and $y + 1$ are divisible by $3$.



Can you solve for the smallest such $x$ and $y$?






share|cite|improve this answer





















  • thanx :) for the help
    – Shukraditya Bose
    Jul 29 at 13:20














up vote
2
down vote













Restating @Lord Shark's hint, we have that $3c$ must be a perfect square and $5c$ a perfect cube.



The smallest such number $c$ would obviously have prime factors of the form $3^x5^y$.



Now, $3^x+15^y$ is a perfect square $implies$ $x + 1$ and $y$ are even.



And, $3^x5^y+1$ is a perfect cube $implies$ $x$ and $y + 1$ are divisible by $3$.



Can you solve for the smallest such $x$ and $y$?






share|cite|improve this answer





















  • thanx :) for the help
    – Shukraditya Bose
    Jul 29 at 13:20












up vote
2
down vote










up vote
2
down vote









Restating @Lord Shark's hint, we have that $3c$ must be a perfect square and $5c$ a perfect cube.



The smallest such number $c$ would obviously have prime factors of the form $3^x5^y$.



Now, $3^x+15^y$ is a perfect square $implies$ $x + 1$ and $y$ are even.



And, $3^x5^y+1$ is a perfect cube $implies$ $x$ and $y + 1$ are divisible by $3$.



Can you solve for the smallest such $x$ and $y$?






share|cite|improve this answer













Restating @Lord Shark's hint, we have that $3c$ must be a perfect square and $5c$ a perfect cube.



The smallest such number $c$ would obviously have prime factors of the form $3^x5^y$.



Now, $3^x+15^y$ is a perfect square $implies$ $x + 1$ and $y$ are even.



And, $3^x5^y+1$ is a perfect cube $implies$ $x$ and $y + 1$ are divisible by $3$.



Can you solve for the smallest such $x$ and $y$?







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 29 at 8:04









iamwhoiam

1,006612




1,006612











  • thanx :) for the help
    – Shukraditya Bose
    Jul 29 at 13:20
















  • thanx :) for the help
    – Shukraditya Bose
    Jul 29 at 13:20















thanx :) for the help
– Shukraditya Bose
Jul 29 at 13:20




thanx :) for the help
– Shukraditya Bose
Jul 29 at 13:20












 

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