Cubes and squares

Question

Consider there are five consecutive numbers $a,b,c,d$ and $e$ such that $a \lt b \lt c \lt d\lt e$.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. :)

So $b+c+d=3c$ and $a+b+c+d+e=5c$? – Angina Seng Jul 29 '18 at 07:33 — Angina Seng, Jul 29 '18 at 07:33

score 2 · Answer 1 · answered Jul 29 '18 at 08:04

2

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, $3^{x+1}5^y$ 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 $y$?

answered Jul 29 '18 at 08:04

iamwhoiam

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thanx :) for the help – Shukraditya Bose Jul 29 '18 at 13:20

Cubes and squares

1 Answers1