v_Enhance wrote: Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime. Evan o'Dorney The title and the post contradict each other

If $a^2-bc=k^2$ then $(a-k)(a+k)=bc$. Let $\gcd(a-k, b)=x$, say $b=xy$ and $a-k=xz$ with $\gcd(y, z)=1$. Then $xz(a+k)=xyc$, so $(a+k)z=cy$. The left side is divisible by $z$, so the right side must be too. Hence $c$ must be divisible by $z$ because $y$ and $z$ have no common factors. Hence $c=wz$ for some $w$; then $(a+k)z=wyz$ and $a+k=wy$. Then $2a+b+c=(a+k)+(a-k)+b+c=wy+xz+xy+wz=(w+x)(y+z)$ is not prime, because $w, x, y, z\geq1$.

my soloution: $(2a+b+c)(2a-b-c)=4a^2-b^2-c^2-2bc=4x^2-(b-c)^2$ let assume that $2a+b+c=p $ such that $p$ is aprime number then$ p(2a-b-c)=(2x-b+c)(2x-c+b) $ $ p | (2x-b+c)(2x-c+b) $ so $P$ divides one of them .let $p | (2x-b+c) $ $ a^2-bc=x^2 $ so $x<a $ then $p=2a+b+c<= 2x-b+c< 2a-b+c $ that is an contradiction. and we are done

Let $ a^2-bc=k^2$ Hence $ (a-k)(a+k)=bc$ ,further work yields that $( a-k+b)/b=(a+k+c)/(a+k),(a-k+b)/(a+k+c)=b/(a+k) $ adding 1 to both sides and simplifying yields $ 2a+b+c=(b+a+k)(c+a+k)/(a+k)$ let gcd$(a+k,c)=r $ and $ a+k=x.r$ and $ c=y.r $ and gcd$(x,y)=1 (b+a+k)(c+a+k)/(a+k)=(b+a+k)(xr+yr)/xr=(b+a+k)(x+y)/x $ obviously gcd$(x+y,y)=1 $ therefore $x|b+a+k $hence$ 2a +b+c$ has two factors $ x+y >1 $ and $ (b+a+k)/x>1(a+k=x.r and r>=1)$

We use the factoring lemma. Let $a^2-bc=x^2$ and rearrange as $(a-x)(a+x)=bc$. Thus $\exists m,n,p,q\in \mathbb{Z}^+:$ $a-x=mn$, $a+x=pq$, $b=mp$, $c=nq$. Solving yields $a=\frac{mn+pq}{2}$, $x=\frac{pq-mn}{2}$. Then $2a+b+c$ $=mn+pq+mp+nq$ $=m(n+p)+q(n+p)$ $=(m+q)(n+p)$.

v_Enhance wrote: Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime. Evan o'Dorney Suppose that $a^2-bc=z^2$ , so $a>z$ so $2a+b+c=\dfrac{(a+c-z)(a+c+z}{c}$ so no prime!

General if $ab=cd$ for positive integers $a,b,c,d$ then $a+b+c+d$ is not prime. Now $a^2-bc=m^2$ , $(a-m)(a+m)=bc$ so $(a-m)+(a+m)+b+c=2a+b+c$ is not a prime.

Let's post a new solution that aren't just a rehash of the first two, shall we? Given the polynomial $bx^2+2ax+c$ we know it has rational roots since its discriminant is a square. So we can factor it in $\mathbb{Q}$. By Gauss's lemma, we can factor it in $\mathbb{Z}$ and WLOG the leading coefficients of both factors are positive. Then since by Descartes' rule of signs this quadratic has only negative roots, all four coefficients of the linear factors are positive. Then we know that plugging in 1 gives two factors which are greater than 1. This implies our conclusion.

The only reason I posted a solution is to state the factoring lemma, which no one before me stated for some reason. IMO, the factoring lemma is at least as useful as say LTE, but I secretly believe that it holds the key to solving Diophantines. :p

I knew all about the factoring lemma, and I used it in my first post. I just didn't call it that, because I think it's too general of a name, and I would rather prove it and have the post be self-contained than cite some theorem and force the readers to either go look it up, trust me or try to prove it themselves, especially since the proof is so easy to mold into the context of the problem.

You didn't state it, which was my concern. I think too many people prove it without emphasizing that it is a useful lemma in its own right, not just for this or that particular problem. As I said, I think it should be given its own name, because it is extremely useful for solving Diophantines.

Let $a^2 - bc = d^2$. Suppose that $2a + b + c$ equals a prime $p$. The $c = p - 2a - b$ Plugging into the first equation, we get $a^2 - b(p - 2a - b) = d^2$ $=>$ $(a+b)^2 - bp = d^2$ $=>$ $(a+b)^2 - d^2 = bp$ $=>$$(a+b+d)(a+b-d) = bp$ One can easily prove $p$ is greater than each of the two factors on the right using the fact that $p = 2a + b + c$ and $a > d$, contradicting the fact that $p$ divides one of them.

This lemma kills the problem: If $a,b,c,d\in N$ and $ab=cd$ Then there exist 4 natural numbers $p,q,r,s$ such that: $a=pq,b=rs,c=pr,d=qs$

Let $a^2 -bc =q^2$. Then $a^2 +2ab +b^2 -bc -2ab-b^2 = q^2 \Longleftrightarrow (a+b)^2 -q^2 = b^2 + 2ab + bc \Longleftrightarrow (a+b-q)(a+b+q)= b(2a+b+c).$ But $q^2 =a^2 - bc < a^2 \Longrightarrow q +1\leq a$. Hence $\frac{a+b-q}{b}= \frac{a-q}{b} + 1 >1$. Done.

solving this by induction as $a^2 -bc<a^2$ , consider it to be equal to $(a-1)^2$ $\implies$ 1+bc=2a putting this in 2a+b+c we get (1+b)(1+c) which is composite now consider $a^2 -bc=(a-n)^2$ we get, $a^2-bc=a^2 +n^2 -2an $$\implies$$ $n^2+bc=2an $\implies$ $2a=\frac{bc+n^2}{n}$ as 2a positive integer $\implies$ $\frac{bc+n^2}{n}$ is positive integer again put value of 2a in $2a+b+c$ $\implies$ $\frac{bc+n^2}{n}+b+c$ $\implies$ $\frac{bc+n^2+nb+nc}{n}$ $\implies$ $\frac{(n+b)(n+c)}{n}$ as n+b>n and n+c>n ,also n|(n+b)(n+c) hence proved that 2a+b+c is composite

Αρχιμήδης 6 wrote: General if $ab=cd$ for positive integers $a,b,c,d$ then $a+b+c+d$ is not prime. Now $a^2-bc=m^2$ , $(a-m)(a+m)=bc$ so $(a-m)+(a+m)+b+c=2a+b+c$ is not a prime. It's a Baltic Way task https://artofproblemsolving.com/community/c6h220674_abcd_is_not_prime_if_abcd

sco0orpiOn wrote: my soloution: $(2a+b+c)(2a-b-c)=4a^2-b^2-c^2-2bc=4x^2-(b-c)^2$ let assume that $2a+b+c=p $ such that $p$ is aprime number then$ p(2a-b-c)=(2x-b+c)(2x-c+b) $ $ p | (2x-b+c)(2x-c+b) $ so $P$ divides one of them .let $p | (2x-b+c) $ $ a^2-bc=x^2 $ so $x<a $ then $p=2a+b+c<= 2x-b+c< 2a-b+c $ that is an contradiction. and we are done **(2a+b+c)(2a-b-c)=4a^2-(b+c)^2

It is 1:45 am and I have PSAT tomorrow. Better do this quick Lemma: if $w,x,y,z$ are positive integers such that $wx=yz$ then $w+x+y+z$ is not prime Proof: Suppose for the sake of contra $w+x+y+z=p$ for some prime $p$. Recall $wx=yz$ so let $w=\frac{yz}{x}$ $x+y+z+\frac{yz}{x}=p$ $=\frac{x^2+xy+zx+yz}{x}$ $=\frac{(x+y)(x+z)}{x}=p$ so, $p|(x+y)(x+z)$. Recall $p$ is prime so $p$ divides $x+y$ or $x+z$. WLOG let $p$ divide $x+y$. This implies $p \leq x+y$ so, $p<w+x+y+z$. Contradiction. Back to the problem. Let $a^2-bc=d^2$. Clearly, $a>d$. $a^2-bc=d^2$ $\Leftrightarrow (a+d)(a-d)=bc$ $a+b$, $a-d$, $b$, and $c$ are all positive integers and $(a+d)(a-d)=bc$ so, $a+d+a-d+b+c=2a+b+c$ is not prime.

v_Enhance wrote: Let $a,b,c$ be positive integers such that $a^2 - bc$ is a square. Prove that $2a + b + c$ is not prime. Evan o'Dorney Suppose we've $a^2 -bc =l^2$ for some $l \in \mathbb{Z}$ Then we can write this as follows $$a^2 +b^2 +2ab-l^2 =bc+2ab+b^2 \Leftrightarrow (a+b)^2 -l^2 =b(2a+b+c) \Leftrightarrow (2a+b+c)= \frac{(a+b+l)(a+b-l)}{b}=\left(1+\frac{a+l}{b} \right)\left(1+\frac{a-l}{b} \right)$$Now as $a,b,c \in \mathbb{Z}^{+}$ we've $a^2 -l^2 >0 \Leftrightarrow (a-l)(a+l) >0 \Leftrightarrow a >l$ therefore it follows that the factors of $(2a+b+c)$ are $>1$ from which it follows that $(2a+b+c)$ is composite $\blacksquare$

My solution: Suppose for the sake of contradiction that $2a + b + c$ is a prime $p$. Let $a^2 - bc = d^2$ for some positive integer $d$. Note that $d < a$. Now as $p = 2a + b + c$ we have $2a \equiv -b-c \pmod p$. Squaring yields $4a^2 \equiv (b+c)^2 \pmod p$. Using the fact $a^2 - bc = d^2$ the above congruence implies $4d^2 + 4bc \equiv (b+c)^2 \pmod p$ which gives $4d^2 \equiv (b-c)^2 \pmod p$ and thus $p \mid |2d - b + c|\cdot|2d + b - c|$ and so $p$ divides one of $|2d - b + c|$ or $|2d + b - c|$. However $|2d - b + c|, |2d + b - c| < |2d + b + c| = 2d + b + c < 2a + b + c = p$, a clear contradiction. Hence $2a + b + c$ must be composite.

For storage. We are given: $$a^2 - bc=x^2,$$where $a,b,c,x\in\mathbb N$. Since equation is symmetric wrt $b,c$, wlog assume $b\geq c$. Also, $a,b,c,x$ are positive, therefore $a^2> a^2 - bc=x^2\implies a>x$. Assume for the sake of contradiction that $p=2a+b+c$, where $p\in\mathbb P$, thus we have $a=\frac{p-b-c}{2}$ and let us substitute that in and we obtain the following: $$\left(\frac{p-b-c}{2}\right)^2 - bc=x^2\Longleftrightarrow p^2-2p(b+c)+(b-c)^2-(2x)^2=0.$$Taking modulo $p$, we obtain $p\mid (b-c-2x)(b-c+2x)$ and since $p\in\mathbb P$, we must have $p\mid (b-c-2x)$ or $p\mid (b-c+2x)$. Hence, we must have $p\leq \vert b-c-2x\vert $ or $p\leq \vert b-c+2x\vert $. We examine each case separately: $\bullet$ $2a+b+c\leq b-c-2x \Longleftrightarrow 2a+2x+2c\leq 0$, clear contradiction. $\bullet$ $2a+b+c\leq 2x+c-b \Longleftrightarrow 2a+2b\leq 2x$, contradiction, since $a>x$. $\bullet$ $2a+b+c\leq b-c+2x\Longleftrightarrow 2a+2c\leq 2x$, contradiction, since $a>x$. $\bullet$ $2a+b+c\leq c-b-2x\Longleftrightarrow 2a+2x+2b\leq 0$, clear contradiction. We are done.

Let $a^2-bc=k^2$. Then $(a-k)(a+k)=bc$. Let $s=a-k,t=a+k$. We need to show that $st=bc$ implies $s+t+b+c$ is not prime. Suppose it was a prime $p$. Observe that modulo $p$, the polynomials $x^2+(s+t)x+st$ and $x^2-(b+c)x+bc$ are the same, hence $b,c$ are $-s,-t$ in some order modulo $p$. But this is a contradiction because $s,t,b,c$ are all positive so this would imply $s+t+b+c\ge 2p$.

Assume that $a^2 -bc =t^2$ Then $a^2 +b^2 +2ab-t^2 =bc+2ab+b^2 \Leftrightarrow (a+b)^2 -t^2 =b(2a+b+c) \Leftrightarrow (2a+b+c)= \frac{(a+b+t)(a+b-t)}{b}=\left(1+\frac{a+t}{b} \right)\left(1+\frac{a-t}{b} \right)$ and since $a^2-t^2 >0 \Leftrightarrow (a-t)(a+t) >0 \Leftrightarrow a >t$ both the factors are $>1$ and we are done.

Posting my solution merely because I think this formulation of the idea has not been posted before. Assume FTSOC that $2a+b+c = p$ is a prime. Notice that $$(a-k)^2 = a^2-bc$$meaning that $$k(2a-k) = bc$$for some $0 < k < 2a-1$. Then notice that $$bc \equiv k(2a-k) \equiv k(-b-c-k) \pmod{p}$$and therefore $$p \mid (b+k)(c+k)$$which is not possible because $$0 < b+k,c+k < 2a+b+c = p$$

Posting for storage FTSOC $a^2-bc=n^2$ and $2a+b+c=p$ $$c=p-2a-b \implies a^2-b(p-2a-b)=n^2 \implies (a+b)^2-n^2=bp \implies (a+b+n)(a+b-n)=bp $$Now note that $a+b+n > b \implies 0<a+b-n<p$ but because $p$ is a prime $p|(a+b+n) \implies p \leq (a+b+n)$ $\implies b \geq a+b-n$ which is clearly a contradiction since $a>n$

Say $a^2-bc=k^2$ for some natural number $k$. FTSOC assume $2a+b+c=p$ where $p$ is some prime. Then note that $$a^2-bc=a^2-b(p-2a-b)=a^2-bp+2ab+b^2=(a+b)^2-bp=k^2$$Therefore $$(a+b-k)(a+b+k)=bp$$Now clearly, $p>b$. Therefore $\nu_p{(b \cdot p)}=1$. Hence, $p$ must divide exactly one of the factors in the LHS. If $p \mid a+b-k$, we see that $a+b+k \leq c <p \leq a+b-k$, contradiction. Therefore $p \mid a+b+k$. Now note that $$p \mid a+b+k \implies k \geq (2a+b+c)-(a+b)=a+c \implies k^2=a^2-bc \geq a^2+c^2+2ac \implies -c^2-2ac-bc \geq 0$$which is absurd $\blacksquare$

Let $a^2-bc=x^2$. Equivalently, $(a-x)(a+x)=bc$. Note that $a-x$ is positive. By Four-Numbers theorem, there exist positive integers $p,q,r,s$ with $a-x=pq$, $a+x=rs$, $b=pr$ and $c=qs$. Hence $$2a+b+c=(a-x)+(a+x)+b+c=pq+rs+pr+qs=(p+s)(q+r)$$ which is composite.

Let $a^2-bc=k^2$, then note that \[2a+b+c = 2a+b+ \frac{a^2-k^2}{b} = \frac{2ab+b^2+a^2-k^2}{b} = \frac{(a+b)^2-k^2}{b} = \frac{(a+b-k)(a+b+k)}{b}\]But note that $b\mid a^2-k^2$, so we can find some pair $b_1,b_2$ such that $b_1b_2=b$ and $b_1\mid a-k, b_2\mid a+k$, so \[2a+b+c = \frac{a+b-k}{b_1}\frac{a+b+k}{b_2} = \left(b_2 + \frac{a-k}{b_1}\right) \cdot \left(b_1+\frac{a+k}{b_2}\right)\]which is clearly composite and we're done. $\blacksquare$.

7 minutes Let $a^2-bc=k^2$. Then $bc=a^2-k^2=(a+k)(a-k)$. Now we also have \[2a+b+c=2a+b+\frac{(a-k)(a+k)}{b}=\frac{a^2-k^2+2ab+b^2}{b}=\frac{(a+b-k)(a+b+k)}{b}\] Note that $b\mid (a-k)(a+k)$. We can let $x$ and $y$ satisfy $x\mid a-k$ and $y\mid a+k$, and $xy=b$. So $\frac{a+b-k}{x}$ and $\frac{a+b+k}{y}$ are both positive integers greater than $1$ that multiply to $\frac{(a+b-k)(a+b+k)}{b}$, so it is not a prime. $\blacksquare$

Let the square be $d^2$. $a^2-bc=d^2$ $a^2-d^2=bc$ $(a+d)(a-d)=bc$ FTSOC, let $2a+b+c$ equal a prime $p$. $2ac+bc+c^2=pc$ $2ac+a^2+c^2-d^2=pc$ $(a+c)^2-d^2=pc$ $(a+c+d)(a+c-d)=pc$ Since $a-d>0$ from line 2, $a>d$. Therefore, $a+c+d$ and $a+c-d$ are greater than $c$. Now, $p$ is greater than $a+c+d$, $a+c-d$, and $c$. Due to that inequality, $\nu_p(LHS)=0 \neq \nu_p(RHS)=1$. Contradiction, $2a+b+c$ cannot be prime.

Here’s an interesting solution which is essentially the same as others. Consider an integer polynomial $P(x)=bx^2+2ax+c$. The root of $P(x)$ will be $-a\pm \sqrt{a^2-bc}$ which are distinct integer roots (unless $a^2=bc$ which can be easily dealt with). But we also have that $P(1)=2a+b+c$. After this, we can use the fact that $P(k)=0,P(l)=0,P(1)=2a+b+c$ and $x-y\mid P(x)-P(y)$ to finish the problem.

Suppose that $a^2-bc=m^2$, so $$(a+m)(a-m)=bc.$$Let $s=a+m$ and $d=a-m$< so $$sd=bc,$$and we wish to show that $s+d+b+c$ is not prime under the condition that $s+d$ is even. Let $K=sd=bc$. Then, we will do cases. Case 1: $K$ is odd. Then $s,d,b,c$ are all odd, so clearly $s+d+b+c$ is not prime. Case 2: $K$ is even. Then, at least one of $s$ and $d$ is even, so both are even since their sum is even. Then, if $b,c$ are both even, then $s+d+b+c$ is even and not prime, so consider the case where WLOG $c$ is odd. Then, $b$ would have to be a multiple of 4 since both $s$ and $d$ are even. Substitute $s=2x,d=2y,b=4z$ so that $$xy=zc$$and we wish to show that $2x+2y+4z+c$ is not prime. Rewrite $y=\frac{K}{x}$ and $z=\frac{K}{c}$ so that $$2x+2y+4z+c$$$$=2x+\frac{2K}{x}+c+\frac{4K}{c}$$$$=\frac{2x^2c+2Kc+xc^2+4Kx}{xc}$$$$=\frac{(2x+c)(xc+2K)}{xc}.$$Let $f=\gcd(x,c)$ and write $x=nf$ and $c=vf.$ Note that by definition, we have both $x\mid K$ and $c\mid K$, which means that $nvf\mid K$. Our expression now becomes $$\frac{(2nf+vf)(nvf^2+2K)}{nvf^2}$$$$\frac{(2n+v)(nvf^2+2K)}{nvf}.$$Note that since $K$ is divisible by $nvf$, we have $nvf^2+2K$ is divisible by $nvf$, and furthermore $nvf^2+2K>nvf$, so $$\frac{nvf^2+2K}{nvf}$$is an integer greater than 1. Since it is being multiplied by another integer greater than 1, it is not prime and we are done. remark: variables used: $a,b,c$ $m$ $s,d$ $K$ $x,y,z$ $f$ $n,v$

Let $a^2-bc=d^2,$ for a integer $d.$ So, $bc=a^2-d^2,$ so $2a+b+c=2a+b+\frac{a^2-d^2}{b}=\frac{a^2-d^2+2ab+b^2}{b}=\frac{(a+b+d)(a+b-d)}{b},$ also note that $b,$ is a divisor of either $a+d,$ or $a-d.$ Now, choose $2,$ numbers that are divisors by $a-d, a+d$, and that their product is $b.$ Then the quotient of these $2$ numbers with $(a+b+d), (a+b-d),$ are both positive integers, that are not $1,$ hence, $2a+b+c,$ is not prime. this took me 25 min (I think to easy for ELMO)

We have $a^2-bc=d^2$. Suppose for contradiction $2a+b+c=p$. Clearly, $p$ is odd as $p>2$. Note that $4d^2=4a^2-4bc\equiv (b+c)^2-4bc=(b-c)^2\pmod{p}$, which means $2d\equiv\pm(b-c)\pmod{p}$. WLOG assume $2d\equiv b-c\pmod{p}$. Then, we have $|2a+b+c|=|p|\leq|2d+c-b|$. Suppose $2d+c\leq b$, then $2a+b+c\leq b-2d-c\Longrightarrow 2(a+d+c)\leq0$, which is impossible. Hence, $2d+c>b$, which means $2a+b+c\leq 2d+c-b\Longrightarrow a<a+b\leq d$, while $a^2>d^2\Longrightarrow a>d$. Hence, we have a contradiction. $\blacksquare$