Is a^2+c Always a Prime Number Under Certain Conditions?

[shivakumar06]

dear sir, i wish to know if i am correct. a^2+c can be a prime number provided if a is even then c is odd or vice versa, also a and c are not multiple of same number. and c is not a negative square of any number. finally prime number is unique combination of 1,2,and other powers of 2. each power of two is used only once. i wish to know if prime number is bound by it
thank you

 

[Vanadium 50]

I don't think you have shown more than those are odd numbers. Showing that a number can be prime, because it's odd, is not terribly useful. For a = 3 and b = 2, you getg 11, which is prime. For a = 5 and b = 2 you get 27, which is not.

I don't understand your last two sentences.

 

[fresh_42]

finally prime number is unique combination of 1,2,and other powers of 2

Every integer can be expressed this way. It's the binary or 2-adic representation. (I assume you are talking about integers, although it is not quite clear to me, since you just say "numbers" and "negative squares".)

each power of two is used only once. i wish to know if prime number is bound by it
thank you

If you mean "once and only once" by "only once" it is wrong for $1+2+4+8 = 15$.
If you mean "at most once" by "only once" it is wrong since all integers have such a representation.
So the final answer to your question seems to be: No.

 

[shivakumar06]

I don't think you have shown more than those are odd numbers. Showing that a number can be prime, because it's odd, is not terribly useful. For a = 3 and b = 2, you getg 11, which is prime. For a = 5 and b = 2 you get 27, which is not.

I don't understand your last two sentences.

sir if we add a condition that a^2-c should not be divisible a-c. will this satisfy condition for prime number?

 

[fresh_42]

What about $a = 12$ , $c = 25$ ? $a$ is even, $c$ is odd, they don 't have a common divisor, $a - c = -13$ does not divide $a^2 -c = 144 - 25 = 119 = 7 * 17$ and $a^2 + c = 144 + 25 = 13^2$ is not prime.