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Suppose, $a$ and $b$ are two positive Integers such that there exist two consecutive positive integers $c$ and $d$ where $$a-b=a^2c-b^2d$$ We have to prove that $|a-b|$ is a perfect square.

I tried to approach the problem in this way:

$$a-b=a^2c-b^2(c+1)\quad( \text{assuming}\;c<d)$$ $$OR,\quad b^2=(a-b)\,[c(a+b)-1]$$

Now, how do I prove that $(a-b)$ and $c(a+b)-1$ are co-prime?

Because if $ab$ is a perfect square and $(a,b)=1$, then both $a$ and $b$ are perfect squares.

Bill Dubuque
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Rijhi
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  • We need only that all $3$ are coprime, i.e. $,(\color{#c00}b,a!-!\color{#c00}b,c(a!+!\color{#c00}b)-1) = (b,\color{#0a0}a,c\color{#0a0}a!-!1) = (b,a,-1) = 1,,$ thus $,b^2=(a!-!b),[c(a!+!b)!-!1]\Rightarrow |a-b|$ is a square, by the Lemma in the dupe. This proof using gcds works more generally than proofs using prime factorization (and is more constructive). – Bill Dubuque Apr 22 '22 at 08:14
  • In the above gcd calculation we used gcd mod reduction, i.e. we can reduce all other gcd args mod one fixed arg without changing the gcd, e.g. in the first equality we reduced the 2nd & 3rd args mod the 1st arg $,\color{#c00}b,,$ and in the next equality we reduced the 3rd arg mod the 2nd $,\color{#0a0}a,,$ i.e. $,(c\color{#0a0}a-1)\bmod \color{#0a0}a, =, -1.,$ See here for elaboration. – Bill Dubuque Apr 22 '22 at 08:32

2 Answers2

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Let $p \in \mathbb{P}$ be a prime number such that $p|b^2$.

Then the following are simultaneously true: \begin{cases} p|b \\ p|(a-b) \quad \text{or} \quad p| c(a+b)-1 \end{cases}

Now if $p|a-b$ then since $p|b$, $p$ is also divided by $a$.

Hence, $p|c(a+b)$, and thus $p \nmid c(a+b) -1$.

Thus, since $v_p(b^2)$ is even, $v_p(a-b)$ is even.

Thus $\mid a-b \mid$ is a square. $\square$

Note If you are not quite familiar with the notation, $v_p(x)$ is the power of prime $p$ in the prime representation of $x \in \mathbb{Z+}$. But that is quite unnecessary in this case. You can deduce that $\gcd(a-b, c(a+b)-1)=1$ anyways.

Nikola Tolzsek
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0

You are on the right track.

If a prime $p$ divides $a-b$, then since $$ b^2=(a-b)\,[c(a+b)-1] \,,$$ we deduce that $p$ also divides $b$. Thus $p$ divides $a+b=a-b+2b$, so it cannot divide $ c(a+b)-1 $.

Yuval Peres
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