Finding lowest possible value for conditional set of variables.

Question

I came across the following problem and am having trouble solving it. I believe I have come across the first insight into how to solve the problem but I am having trouble seeing the second insight.

The problem is as follows:

Suppose you have 5 positive sequential integers $$ r, o, h, a, n. $$ The sum of $$ o + h + a $$ is a perfect square. The sum of $$ r + o + h + a + n $$ is a perfect cube. What is the lowest possible value of h?

So my first insight is as follows:

$$ r + o + h + a + n = r + (r+1) + (r+2) + (r+3) + (r+4) = 5r+10 $$ $$ \implies 3r+6=y^2 $$ $$ and, 5r+10=x^3. $$

I am unsure how to combine these functions or use them to find a minimum, since they do not share an x or y.

Any hints/suggestions are appreciated.

EDIT:

I made an error in transposing the problem. The variables are not defined to be "sequential positive integers", but as "consecutive positive integers". I am unsure if this implies a variable range between the numbers or not.

If it does, I get the answer of:

$$ r=1, o=2, h=3, a=4, n=17 $$

I cannot imagine a lesser solution since r is the lowest possible positive integer and o and h increase by 1 each time. However, my answer is not correct. Does "consecutive positive integers" imply a uniform range between each number?

Yes, consecutive means that any two neighbors differ by 1. So 1,2,3 are consecutive integers, but 1,3,4 are not consecutive. — ChocolateAndCheese, Sep 29 '15 at 23:42
This is a buttinsky as it doesn't change anything. rather than doing things in terms of r, do them in terms of a h. That way when you say o+h+a = 3r + 6, you would simply have have o + h + a = 3h, and instead of r + o + h + a +n = 5r + 10, you'd simply have r + o + h + a +n = 5h. It won't change anything but it makes the arithmetic less irritating. — fleablood, Sep 30 '15 at 00:25

ChocolateAndCheese · Answer 1 · 2015-09-30T00:13:29.310

2

Your first two insights are good. Let's keep going, though. Rearrange your equations $3r +6 = y^2$ and $5r + 10 = x^3$ to get $r = \frac{1}{3}(y^2 - 6)$ and $r = \frac{1}{5}(x^3 - 10)$. Then we get \begin{equation} \frac{y^2}{3} - 2 = \frac{x^3}{5} - 2 \end{equation} or \begin{equation} \frac{y^2}{3}= \frac{x^3}{5}. \end{equation} I believe that the lowest integer solution to this occurs when $x=15$ and $y=45$ (thank you to @joriki for pointing that out!). We can then find $r = 673$, and then $h = 675$.

edited Sep 30 '15 at 00:13

answered Sep 29 '15 at 23:22

ChocolateAndCheese

1,542

Okay, thank you for that confirmation. However, I apparently did not transpose the problem correctly! Please see my edit. – d0rmLife Sep 29 '15 at 23:40
How about $x=15$, $y=45$, $r=673$? – joriki Sep 30 '15 at 00:03
1

@joriki Thanks. You're completely correct. – ChocolateAndCheese Sep 30 '15 at 00:14
@joriki Did you find your answer using an approach similar to the one described by lulu? If not, what was your approach? – d0rmLife Sep 30 '15 at 04:47
@d0rmLife: Somewhat similar, but less formalized -- I just multiplied through by $3$ and $5$ and then kept adding factors of $3$ or $5$ to $x$ or $y$ as long as one side had too few of them. – joriki Sep 30 '15 at 06:00

lulu · Accepted Answer · 2015-09-30T00:30:25.980

2

It is easier, I think, to write the sequence as $$\{h-2,h-1,h,h+1,h+2\}$$

Then the conditions are $$3h=n^2\;\;and\;\;5h=m^3$$

Clearly $ord_3(h)$ must be an odd multiple of $3$, and $ord_5(h)$ must be even and congruent to $-1\,mod(3)$. (we also know that $6|ord_p(h)$ for all other primes, but for the minimal $h$ we will just zero all of those out). It follows that the smallest suitable $h$ is $h=3^35^2=675$.

For avoidance of doubt: In the above $ord_p(h)$ refers simply to the greatest power of $p$ dividing $h$. Thanks to a commenter (@joriki) for pointing out another (unintended) use of that notation.

edited Sep 30 '15 at 00:30

answered Sep 30 '15 at 00:17

lulu

70,402

It seems you're using $\operatorname{ord}_n(h)$ with a different meaning from this convention? – joriki Sep 30 '15 at 00:23
Indeed I am! When did that change? For me, $ord_p(n)$ means the highest power of $p$ which divides $n$. How does one refer to that these days? – lulu Sep 30 '15 at 00:26
1

@joriki I have some support, e.g. http://math.stackexchange.com/questions/116356/what-is-the-meaning-of-operatornameord-pn – lulu Sep 30 '15 at 00:30
I see. We should tell Wikipedia about that :-) – joriki Sep 30 '15 at 01:14

score 0 · Answer 3 · edited Oct 05 '15 at 18:26

Cute puzzle!

$o + h + a = h - 1 + h + h + 1 = 3h = v^2$.

So $3$ divides $v$.

$h = 3\times v\times\dfrac v3$

so $3$ divides $h$.

Similarly $r + o + h + a + n = 5h = w^3$.

so $5$ divides $w$.

so $h = 25\times w\times\left(\dfrac w5\right)^2$.

so $25$ divides $h$.

so $75=3\times25$ divides $h$.

but if $3$ divides $h$ then $5\times3\times\frac h3 = w^3$. so $3$ divides $w$. so

$5h = 27\left(\dfrac w3\right)^3$.

So $27$ divides $h$ and $25$ divides $h$ so $h = 27\times25$ is the lowest possible $h$. That is, if $h = 27\times25$ is a solution. If so:

$o + h + a = 3h = 3\times27\times25 = (9\times5)^2$

$r + o + h + a + n = 5h = 5\times27\times25 = (3\times5)^5$.

So that's it. $r = 27\times25 - 2, o = 27\times25 -1, a = 27\times25 + 1, n = 27\times25 + 2$, etc.

$3h=v^2$ and $h=3\cdot v\cdot(v/3)$ don't fit together. – joriki Sep 30 '15 at 01:16 — joriki, Sep 30 '15 at 01:16

Finding lowest possible value for conditional set of variables.

3 Answers3