let $a, b, c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

Question

I need help with the following problem:

let $a, b, c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime.

thanks

Answer 1

Given $a^2-bc=n^2$, we get: $$2a+b+c=2a+b+\frac{a^2-n^2}{b}=\frac{2ab+b^2+a^2-n^2}{b}=\frac{(a+b)^2-n^2}{b}=\\ \frac{(a+b-n)(a+b+n)}{b}.$$

Note: $a+b-n>b$ and $a+b+n>b$.

There can be three cases:

Case 1: $a+b-n=bp$, then $2a+b+c$ is divisible by $p>1$ and $a+b+n>1$.

Case 2: $a+b+n=bq$, then $2a+b+c$ is divisble by $q>1$ and $a+b-n>1$.

Case 3: $a+b-n=uv$ and $a+b+n=st$ such that $us=b$, then $2a+b+c$ is divisible by $v>1$ and $t>1$.

Answer 2

Suppose:

$a^2-bc=k^2$ then $(a-k)(a+k)=bc$

If $( b, c)=1 $ and $[(a-k), (a+k)]=1$, then we have:

$a-k=b$

$a+k=c$

From these relations we get:

$b+c=2a$

⇒ $2a +b+c=2a+2a=4a$

Therefore $2a+b+c$ is not prime.

It is better to put this condition
'(a, b, c)=1' in the question.

Answer 3

Suppose $a,b,c$ are positive integers such that \begin{align*} &a^2 - bc = n^2\\[4pt] &2a + b + c = p\\[4pt] \end{align*} where $n$ is a nonnegative integer, and $p$ is prime.

Our goal is to derive a contradiction.

Then $a^2 > n^2$, hence $a > n$.

Also, we have $c=p-2a-b$, hence \begin{align*} &a^2 - bc = n^2\\[4pt] \implies\;&a^2-b(p-2a-b)=n^2\\[4pt] \implies\;&a^2-bp-2ab+b^2=n^2\\[4pt] \implies\;&a^2-2ab+b^2-n^2=bp\\[4pt] \implies\;&(a+b)^2-n^2=bp\\[4pt] \implies\;&(a+b+n)(a+b-n)=bp\\[4pt] \end{align*} From $(a+b+n)(a+b-n)=bp$ and $a+b+n > b$, we get $0 < a+b-n < p$.

Since $0 < a+b-n < p$, and $p$ is prime, it follows that $\gcd(p,a+b-n)=1$.

Since $\gcd(p,a+b-n)=1$, and $(a+b+n)(a+b-n)=bp$, we get $p{\,\mid\,}a+b+n$, hence $p\le a+b+n$.

Since $p \le a+b+n$, and $(a+b+n)(a+b-n)=bp$, we get $b \ge a+b - n= b+(a-n) > b$, contradiction.