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Let $n,k$ be two natural numbers, $n\ge k+2$ and $$(n-k-1)(n-k)(n-k+1)(n-k+2)=k(k+1)n(n+1).$$ In the LHS we have product of 4 consecutive numbers and in the RHS we have the product of 2 consecutive numbers on the product of 2 consecutive numbers.

Problem. Prove that $k,k+1,n,n+1$ also are consecutive numbers.

My attempt. All factors are in increasing order. Suppose that $n-k-1=k$, then $n=2k+1$ and on we can reduce the products to $k(k+1)(k+2)(k+3)=k(k+1)n(n+1)$ or $(k+2)(k+3)=n(n+1)=(2k+1)(2k+2)$ We have equal products of two 2 consecutive numbers so it implies that $k+2=2k+1$ and we get a solution $k=1, n=3$.

How to prove that the case $n-k-1 \neq k$ is impossible?

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    $\begingroup$ what has been tried ? $\endgroup$
    – user451844
    Sep 1, 2017 at 16:27
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    $\begingroup$ If $n>k+2$ then $k,k+1,n,n+1$ can't be consecutive numbers. $\endgroup$ Sep 1, 2017 at 16:34
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    $\begingroup$ @ThomasAndrews That's got to be a typo. I'll fix the inequality. $\endgroup$
    – Joffan
    Sep 1, 2017 at 16:45
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    $\begingroup$ I improved the question. $\endgroup$
    – Leox
    Sep 2, 2017 at 6:45

1 Answer 1

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Let $a:=n-k\ (\ge 2)$.

Then, we have $$(a-1)a(a+1)(a+2)=k(k+1)(k+a)(k+a+1)$$ This can be written as $$(2k^2+2ka+2k+a)^2=4a^4+8a^3-3a^2-8a$$

Now, for $a\gt 2$, we have $$(2a^2+2a-2)^2\lt 4a^4+8a^3-3a^2-8a\lt (2a^2+2a-1)^2$$ from which we have that $4a^4+8a^3-3a^2-8a$ cannot be a square number for $a\gt 2$.

So, we have to have $n-k=a=2$.

It follows from this that $k,k+1,n,n+1$ are consecutive numbers.

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    $\begingroup$ How did you get that $(...)^2 = ...$ identity. That is the key. Also, "By the way" is an odd phrase, since that introduces the proof. But, an extremely nice proof. $\endgroup$ Sep 4, 2017 at 6:19
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    $\begingroup$ @marty cohen: I tried to have something of the form $(f(a,k))^2=g(a)$ since I know that considering a square number sometimes helps. I think that it was just try and error after starting to try to have such an identity. I'll remove "By the way", thank you. $\endgroup$
    – mathlove
    Sep 4, 2017 at 6:40
  • $\begingroup$ @mathlove Great solution, thank you! $\endgroup$
    – Leox
    Sep 4, 2017 at 15:14

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