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I was calculating following for some numbers and I found this is true for a lot of examples.

Let $p\geq 3$ be a prime number such that $p|a$ and $p^2|b$ and following equation is true:

$$\frac{a}{p}-\frac{b}{p^2}=-1$$

Now we write $b-a=p_1^{c_1}...p_k^{c_k}$, the prime factorization, then $c_j$s are exactly $1$.

I am not sure this is always true but I took some examples and it worked (probably true for some specific condition on $a$ and $b$, I don't know). I am trying to prove or find counterexample. Just wanted to post here for everyone to know this also. If this is a problem found in some book already, please don't bite me, it came to me when I was trying to sleep.

Bill Dubuque
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MAN-MADE
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1 Answers1

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If $a = 9$ and $b = 36$ then the condition is satisfied with $p=3$ but $b-a = 3^3$.

beeclu
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  • I'm guessing we used the same method: got to $b-a = p^2+pa-a$, and setting $p=3$ and $a=9$ is the simplest combination. – beeclu Mar 25 '25 at 20:07