Prove a sum is a composite number

[anemone]

For positive integers $p,\,q,\,r,\,s$ such that $ps=q^2+qr+r^2$, prove that $p^2+q^2+r^2+s^2$ is a composite number.

 

[MarkFL]

My solution:

Spoiler

From the given:

\(\displaystyle ps=q^2+qr+r^2\)

We obtain:

\(\displaystyle q^2+r^2=2ps-(q+r)^2\)

And so:

\(\displaystyle p^2+q^2+r^2+s^2=p^2+2ps+s^2-(q+r)^2=(p+s)^2-(q+r)^2=(p+q+r+s)(p-q-r+s)\)

Which shows the given expression is composite.

 

[anemone]

Nice one, MarkFL! Thanks for participating too. :)

 

[kaliprasad]

if one talks of algebraic expression being composite I agree but if one talks of the number being composite this is not correct as one of the factors could be 1

 

[Opalg]

if one talks of algebraic expression being composite I agree but if one talks of the number being composite this is not correct as one of the factors could be 1

That possibility is easily eliminated. If $p-q-r+s = 1$ then the equation becomes $p^2+q^2+r^2+s^2 = p+q+r+s$. That can only hold for positive integers if $p=q=r=s=1$. But in that case the equation $ps = q^2+qr+r^2$ does not hold.

 

[kaliprasad]

That possibility is easily eliminated. If $p-q-r+s = 1$ then the equation becomes $p^2+q^2+r^2+s^2 = p+q+r+s$. That can only hold for positive integers if $p=q=r=s=1$. But in that case the equation $ps = q^2+qr+r^2$ does not hold.

addition of above completes the proof. IN other words this was the missing link. Thanks Opalg

 

[anemone]

That possibility is easily eliminated. If $p-q-r+s = 1$ then the equation becomes $p^2+q^2+r^2+s^2 = p+q+r+s$. That can only hold for positive integers if $p=q=r=s=1$. But in that case the equation $ps = q^2+qr+r^2$ does not hold.

Thanks Opalg to the rescue.

Another method (I'll admit that I overlooked the necessity to show that $p-q-r+s \ne 1$, as I thought that was quite obvious and now after pondering about it, I owed MHB a good and solid reason why $p-q-r+s \ne 1$).

Spoiler

We will prove it by contradiction.

Suppose $p+s = 1+q+r$ Then squaring it and using the fact $ps = q^2+qr+r^2$ twice gives

$p^2+s^2+(q-1)^2+(r-1)^2=1$

However, since $p^2+s^2 \ge 2$, we have reached to a contradiction.

Therefore, $p-q-r+s \ge 2$ and $p^2+q^2+r^2+s^2$ is composite.