Prime numbers and variables

Question

What variables can be prime numbers in a following equality: We are working in natural numbers $$b^2=a^2+ca\quad \text {with:}\quad a+b>c, \,c+a>b,\, c+b>a$$

I have managed to prove, that $b$ can not be prime and $c$ can, how to prove that $a$ can/cannot be a prime number. I think it may be connected with a following quadratic equation:

$a^2+c\cdot a-b^2=0$, with roots $a_1,a_2=\frac {-c\pm \sqrt {c^2+4b^2}}2$.

Not sure how to prove it, though

$b$ can be prime if $a=1$, $p^2=1+(p^2-1)\times 1$ for any $p$. — lulu, Apr 25 '21 at 17:55
Yeah, i will rewrite the question, because there are some conditions that do not allow it to be 1 — King Crimson, Apr 25 '21 at 17:56
Please edit to include all the conditions you have in mind...otherwise, people will waste their time (and yours) finding examples that don't meet all your tests. — lulu, Apr 25 '21 at 17:57
Sorry, didn't mean to waste your time, really, my bad, next time i will try to avoid making this mistake. — King Crimson, Apr 25 '21 at 17:57
No worries. I tried to edit edit your post but I did not understand what you meant by $a12$. — lulu, Apr 25 '21 at 17:59
Well, it is an index for roots of quadratic eqation. Something like first root and the second one — King Crimson, Apr 25 '21 at 18:00
Are $a,b,c$ integers? Are thy naturals? We easily have $b=3$ and $a= 2$ and $c=2.5$. — fleablood, Apr 25 '21 at 18:08
$a$ can't be prime. As remarked, if $a=p$ were prime then we'd have $c=kp$. That would mean that $b=\sqrt {1+k}\times p$. But we require $a+b>c$ hence $1+\sqrt {1+k}>k$ which doesn't work. If $k=2$ we have $a+b=c$ and for bigger $k$ we have 1+\sqrt {1+k}<k$. — lulu, Apr 25 '21 at 18:17
@lulu, thank you very much. I had my classes tomorrow(today, actually, at 7 am, but it is already 3 am), but I couldn't come up with a solution within a reasonable time, because i am pretty tired already. You saved me! — King Crimson, Apr 25 '21 at 18:20
@lulu, i will copy your answer to close my question, if you don't mind(will mention you in there). — King Crimson, Apr 25 '21 at 18:25

score 1 · Accepted Answer · answered Apr 25 '21 at 18:29

$a$ can't be prime. As remarked, if $a=p$ were prime then we'd have $c=kp$. That would mean that $b=\sqrt{1+k}×p$. But we require $a+b>c$ hence $1+\sqrt{1+k}>k$ which doesn't work. If $k=2$ we have $a+b=c$ and for bigger $k$ we have $1+\sqrt {1+k}<k$. This answer is not mine, it is lulu's

Prime numbers and variables

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