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I'm working on some equations in number theory and I stuck on below inequality :

$$p^{\alpha +1}q^{\beta +1}-2p^{\alpha+1}q^{\beta}-2p^{\alpha}q^{\beta +1}+2p^{\alpha}q^{\beta}+p^{\alpha+1}+q^{\beta+1}-1 > 0$$

Here $p$ and $q$ are distinct prime numbers and $p,q >2$ and $\alpha\,,\beta$ are positive integer numbers.

Can somebody help me to prove that or find counterexample , although I believe that the inequality is true.

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    $\begingroup$ Are $\alpha,\beta$ real or integer numbers? $\endgroup$
    – Ángel Valencia
    Feb 20, 2016 at 16:48
  • $\begingroup$ @ÁngelValencia: integer number $\endgroup$
    – Mojtaba
    Feb 20, 2016 at 16:50
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    $\begingroup$ Are they positive? Non negative? Or arbitrary integers? Please edit this into the question. $\endgroup$
    – Erick Wong
    Feb 20, 2016 at 16:51
  • $\begingroup$ @ErickWong : sorry , I edited . $\endgroup$
    – Mojtaba
    Feb 20, 2016 at 16:54

3 Answers 3

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Take $p=q=\alpha=\beta=3$. We get $$3^{8}- 4\cdot 3^{7} + 2 \cdot 3^{6} + 3^{4} + 3^{4}-1=-568$$

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  • $\begingroup$ $p$ and $q$ are distinct . $\endgroup$
    – Mojtaba
    Feb 20, 2016 at 17:09
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    $\begingroup$ @A-M: WRONG! Correct is: "Oh, I'm sorry $-$ I forgot to say that $p$ and $q$ are distinct. I will edit my question immediately." $\endgroup$
    – TonyK
    Feb 20, 2016 at 17:15
  • $\begingroup$ @TonyK : Oh, I'm sorry − I forgot to say that p and q are distinct. I will edit my question immediately. $\endgroup$
    – Mojtaba
    Feb 20, 2016 at 17:16
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Factor $p^{\alpha +1}q^{\beta +1}-2p^{\alpha+1}q^{\beta}-2p^{\alpha}q^{\beta +1}+2p^{\alpha}q^{\beta}$ as $$p^\alpha q^\beta(pq-2p-2q+2)=p^\alpha q^\beta\bigl((p-2)(q-2)-2\bigr),$$ hence \begin{align*}&p^{\alpha +1}q^{\beta +1}-2p^{\alpha+1}q^{\beta}-2p^{\alpha}q^{\beta +1}+2p^{\alpha}q^{\beta}+p^{\alpha+1}+q^{\beta+1}\\ {}={}&p^\alpha q^\beta\bigl((p-2)(q-2)-2\bigr)+p^{\alpha+1}+q^{\beta+1}. \end{align*} Now, if $p,q>2$, $p\ne q$, the first term in the above sum is $\ge 0$, and $p^{\alpha+1}+q^{\beta+1}\ge p+q \ge 8 $ since $p,q$ are distinct odd primes.

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    $\begingroup$ That doesn't look right to me: $\cdots+p+q)$ should be $\cdots+\frac{p}{q^\beta}+\frac{q}{p^\alpha})$, shouldn't it? $\endgroup$
    – TonyK
    Feb 20, 2016 at 16:16
  • $\begingroup$ Your factorization is wrong. $\endgroup$
    – Ángel Valencia
    Feb 20, 2016 at 16:48
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    $\begingroup$ @Bernard It is wrong for the reason TonyK says. $\endgroup$
    – wythagoras
    Feb 20, 2016 at 17:05
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    $\begingroup$ @TonyK: That was a silly mistake. I've updated my answer. $\endgroup$
    – Bernard
    Feb 20, 2016 at 17:36
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The LHS is $$x y (\;(p-2)(q-2)-2\;)+p x+q y-1$$ where $x=p^a$ and $y=q^b$. Since $p,q$ are distinct odd positive integers, $(p-2)(q-2)\geq 3.$

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    $\begingroup$ Looks very much like my final answer… $\endgroup$
    – Bernard
    Feb 20, 2016 at 17:46
  • $\begingroup$ @Bernard : Thank you both $\endgroup$
    – Mojtaba
    Feb 20, 2016 at 18:02

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