If we write
a^(2) + b^(2) = c^(2)
&
c^(2) + d^(2) = e^(2)
where "a" is an odd prime then either "c" will be prime or "e" will be prime other wise d/b will become prime but one of the case for prime will occur.
can anybody provide a counterexample of this statement.
2007-01-01 17:53:00 · 9 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
a,b,c,d,e are all perfect squares.
2007-01-01 17:58:08 · update #1
25^(2) + 312^(2) = 313^(2)
313 is a prime number.
2007-01-01 18:24:12 · update #2
no
It is pretty unclear whatg you are brabbling
You should reformulate your question
1) if a is an odd prime then either c is a prime or e is a prime. is understiiood
2) what do you mean with "other wise"
do you mean here that if a is not an odd prime, or do you mean a is not a prime, or do you mean that d/b can be prime ?
Part of Mathematics is that you can formulate clearly your statements, if you do that then you might get an answer.
Ah and an odd prime can not be a perfect square
2007-01-01 21:25:01 · answer #1 · answered by gjmb1960 7 · 2⤊ 0⤋
Counterexample:
a = 7 (an odd prime)
b = 24
c = 25, which is not prime (7^2 + 24^2 = 25^2)
d = 60
e = 65, which is not prime (25^2 + 60^2 = 65^2)
Your question makes a statement about d/b, which I don't understand. In the above example, d/b = 2.5, which of course is not a prime number.
2007-01-02 02:19:42 · answer #2 · answered by actuator 5 · 2⤊ 2⤋
In your details, you declared that a, b, c, d, and so forth is a perfect square. However, you also declared a to be an odd prime. There's no way "a" can be an odd prime *and* a perfect square.
I think you meant to say "a, b, c, d, e, are all integers."
I tried working on your problem and it proved more difficult than I thought. I started by looking at a table of Pythagorean triples, and then had no idea where to go from there.
2007-01-02 02:05:42 · answer #3 · answered by Puggy 7 · 0⤊ 1⤋
Is a^(2) the qwerty equivelant of "a squared"? If not, I completely misunderstood your question. lol
If it IS, then A can't be a prime number like A since C has a square root.
2007-01-02 01:56:37 · answer #4 · answered by Anonymous · 0⤊ 1⤋
A counterexample of this statement might be the Pythagorean Theorem check that link you can find the proofs there.
http://en.wikipedia.org/wiki/Pythagorean_theorem
I hope this help
2007-01-02 02:02:20 · answer #5 · answered by r_dardon 2 · 0⤊ 1⤋
ummm
ok... a=7 b=2 d=2
so, 49+4=c^2
sqrt(53) = c (not prime)
squrt(53)+4 = e^2 (not prime)
none of these are prime because you cant have a prime decimal, because primes are all positive integers. and they are also all odd.
2007-01-02 02:01:56 · answer #6 · answered by cppdungeon 2 · 0⤊ 4⤋
A couple of the answers here make it clear that your problem isn't correctly stated.
2007-01-02 05:10:45 · answer #7 · answered by ninasgramma 7 · 0⤊ 0⤋
that was a good question n i' ve also rated urs as a good question but i m sorry i dont know the proof as i m a weak maths student.sorryyyyyyy!
2007-01-02 01:58:53 · answer #8 · answered by Anonymous · 0⤊ 3⤋
its ur problem u solve urself
2007-01-02 04:37:30 · answer #9 · answered by niki 1 · 0⤊ 1⤋