Show that if p,q are primes, then p^2 + q^2 / p + q, if a whole number, is a prime number as well.
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M$3 Answers
"unless you mean to have parentheses around the p^2 + q^2"
that still wouldn't work for p = q = 2
(2^2 + 2^2) / 2 + 2 = 6 != prime
I think the conjecture is supposed to be:
Given that two numbers, p & q are primes,
(p^2 + q^2)/(p + q) is prime when it is a whole number
if p, q = 2
=> (4 + 4) / (2 + 2) = 2 = prime
However, this only works when p = q, anything else gives a decimal answer (and the way I read the question, decimals are to be excluded totally, not rounded off)
So to simplify, as p = q
2p^2/2p = p^2/p = p, which will always be prime as p is prime to start with (!)
Hope that helps - feel free to re-word your original conjecture.
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M$
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M$2^2 + 2^2 / 2 + 2
4 + 2 + 2 = 8
8 = divisible by 2 & 4 = not prime.
where p & q each = 3
3^2 + 3^2 / 3 + 3
9 + 3 + 3 = 15
again, not prime.
unless you mean to have parentheses around the p^2 + q^2 otherwise those are examples using the order of precedence (multiplication and division prior to addition and subtraction)
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M$
