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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.

$30^3,60^3,90^3,120^3....$

And here is were I got stuck at...

Is the concept I'm using correct?

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  • $\begingroup$ 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
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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 $$ \sqrt{p+q+r+s+t} = \sqrt{5r} = 3\cdot 5^2 = 75. $$

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ $r=1125$ 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
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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 $$\sqrt{p+q+r+s+t}=75.$$

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    $\begingroup$ 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
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    $\begingroup$ Is $q$ the same $q$ in the question? $\endgroup$ – Arc Neoepi Jun 22 '16 at 23:31

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