Five consecutive integers $p,q,r,s,t$,each less than $10000$, produce a sum which is a perfect square,while the sum of $q,r,s$ is a perfect cube.What is the value of $ \sqrt{p+q+r+s+t}$ ?

What I have tried so far:

Let $p=r-2$

$p+q+r+s+t =5r $

$5r=x^2 $

$q+r+s =3r =y^3 $

$x^2 -y^3 =2r $

So,the only perfect squares which are divisible by $5$ are the multiples of $5$: $25,100,225,400...$

I also observed that $100-25=75,225-100=125$,where $125-75=50$.Trying that for $225-100=125,400-225=175$ where $175-125=50$

Then,for the perfect cube which are divisible by 3 and must be less than the perfect squares.


And here is were I got stuck at...

Is the concept I'm using correct?

  • $\begingroup$ I guess to simplify the sum? I have almost no idea how to do these type of questions... $\endgroup$ – Arc Neoepi Jun 22 '16 at 22:37

As $p+q+r+s+t = 5r$ and $q+r+s = 3r$ we have $$ 5r = m^2, \quad 3r = n^3 \tag1 $$ for some integers $m$ and $n$. Thus $5\mid r$ and $3^2\mid r$, so $r = 5\cdot3^2\cdot r_1$ for som integer $r_1$. Substitute it to $(1)$: $$ 5^2\cdot 3^2\cdot r_1 = m^2, \quad 3^3\cdot 5\cdot r_1 = n^3. \tag2 $$ Thus $r_1$ is a perfect square ($r_1 = m_1^2$, where $m_1 = m/15)$ and $5^2\mid r_1$. One may see that $r_1 = 5^2\cdot r_2^2$, where $r_2$ is integer, satisfying bouth these conditions. Substitute it to $(2)$: $$ 5^2\cdot 3^2\cdot 5^2 \cdot r_2^2 = m^2, \quad 3^3 \cdot 5^3 \cdot r_2^2 = n^3. $$ Now we see that $r_2^2$ is a perfect cube. But as $r < 10000$ and $r = 5\cdot 3^2\cdot 5^2\cdot r^2_2$ we get that $r_2^2 \le 8$. There is only one number less than or equal to $8$ which is bouth perfect square and perfect cube - it is $1$. So $r_2 = 1$ and $r = 5\cdot 3^2\cdot 5^2 = 1125$ and $$ \sqrt{p+q+r+s+t} = \sqrt{5r} = 3\cdot 5^2 = 75. $$

  • $\begingroup$ I don't understand from the third paragraph onwards...I'm a bit lost at the 'one way see...' that part $\endgroup$ – Arc Neoepi Jun 22 '16 at 23:11
  • $\begingroup$ @ArcNeoepi $r_1$ is a perfect square and $5^2$ divides $r_1$. So we may represent $r_1$ as $5^2r_2^2$ where $r_2$ is some integer number. This is a perfect square (as $5^2r_2^2 = (5r_2)^2$) and this is divisible by $5^2$ - exactly what we need. $\endgroup$ – Anton Grudkin Jun 22 '16 at 23:16
  • $\begingroup$ Wow,that nice!Tho,I'm gonna need some time to digest this.Thanks! $\endgroup$ – Arc Neoepi Jun 22 '16 at 23:18
  • $\begingroup$ $\frac 5r and \frac 9r $,how does that become $r =45r_1 $? $\endgroup$ – Arc Neoepi Jun 23 '16 at 0:36
  • $\begingroup$ @ArcNeoepi if $5\mid r$ then $r = 5 k$ where $k$ is integer. If also $9\mid r$ we get that $9\mid k$ because $9$ does not divide $5$. Thus $k = 9 r_1$ for some integer $r_1$ and thus $r = 5\cdot 9\cdot r_1 = 45r_1$. This is a consequnce of more general fact, see here: math.stackexchange.com/questions/407540/… $\endgroup$ – Anton Grudkin Jun 23 '16 at 0:39

Since $\frac{3}{5}x^2=y^3$ is a cube, the number of factors of 5 in $x^2$ must be even and one more than a multiple of 3. So choices are $4$ and $10$. Also the number of factors of 3 in $x^2$ must be even and one less than a multiple of $3$, so choices are $2$ and $8$. Now, $\frac{x^2}{5} \le 10000$, so $x$ must be a multiple of $3\cdot 5^2 = 75$. And in fact $x=75$ (so $r = \frac{75^2}{5}=1125$) works.

  • $\begingroup$ $r=1125$ is correct. $\endgroup$ – parsiad Jun 22 '16 at 22:53
  • $\begingroup$ 'the number of factors of $5$ in $x^2$ must be even and one more than a multiple of $3$. So choices are $4$ and $10$.' I don't understand the part about the number of factors part. $\endgroup$ – Arc Neoepi Jun 22 '16 at 23:07
  • $\begingroup$ Since $\frac{3}{5}x^2$ is a cube, $\frac{x^2}{5}$ must be divisible by $5^3$, and $x^2$ has an even number of factors of $5$. Putting these together gives the fact I quoted above. $\endgroup$ – rogerl Jun 23 '16 at 1:06
  • $\begingroup$ Wow,that is really interesting!!!Thanks! $\endgroup$ – Arc Neoepi Jun 23 '16 at 1:09

The problem is that of finding the value of $\sqrt{5p+10}$ where $p$ is an integer less than 9995 such that $$a^2=p+(p+1)+(p+2)+(p+3)+(p+4)=5p+10=5(p+2),$$ $$b^3=(p+1)+(p+2)+(p+3)=3p+6=3(p+2),$$ for some integers $a$ and $b$. Then $5\mid a$ and $3\mid b$, and hence $5\mid p+2$ and $3^2\mid p+2$, meaning that $$p=45n-2,$$ for some integer $n$. It follows that $$a^2=225n=3^2\times5^2\times n\qquad\text{ and }\qquad b^3=135n=3^3\times5\times n.$$ we see that the right-hand sides are squares resp. cubes for $n=5^2$. This yieds $p=1123$ and hence $$a^2=5p+10=75^2\qquad\text{ and }\qquad b^3=3p+6=15^3.$$ Note that the next value of $n$ that works is $n=3^6\times5^2>10000$ and that $n$ cannot be negative, so $$\sqrt{p+q+r+s+t}=75.$$

  • 1
    $\begingroup$ What do you mean when you write $5\mid a$ ? I haven't learned this '$\mid$' yet. $\endgroup$ – Arc Neoepi Jun 22 '16 at 22:53
  • $\begingroup$ By that I simply mean that $5$ divides $a$. $\endgroup$ – Inactive - avoiding CoC Jun 22 '16 at 22:54
  • $\begingroup$ Wouldn't it be $5$ divided by $a^2$ and $3$ divided by $b^3$ and $3$ divided by $p+2$ ? $\endgroup$ – Arc Neoepi Jun 22 '16 at 22:58
  • $\begingroup$ From $a^2=5(p+2)$ it follows that $5$ divides $a^2$. Then also $5^2$ divides $a^2=5(p+2)$, and therefore $5$ divides $p+2$. The same reasoning goes for $b^3=3(p+2)$. $\endgroup$ – Inactive - avoiding CoC Jun 22 '16 at 23:02
  • 1
    $\begingroup$ Is $q$ the same $q$ in the question? $\endgroup$ – Arc Neoepi Jun 22 '16 at 23:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.