Finding Primes!

a a , b b , c c and d d are integers such that 0 < a , b , c , d < 999 0<a,b,c,d<999 and a b = c d ab=cd .

How many ordered sets of integers ( a , b , c , d ) (a,b,c,d) satisfy

a + b + c + d = p a+b+c+d=p

where p p is a prime number?


The answer is 0.

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2 solutions

Jit Ganguly
Jul 12, 2014

C l a i m : Claim: If a , b , c , d a,b,c,d are positive integers satisfying a b = c d ab=cd , then a n + b n + c n + d n a^{n}+b^{n}+c^{n}+d^{n} is composite for any natural n n . P r o o f : Proof: Now, a b = c d ab=cd , then there exists co-prime integers m , k m,k such that a d = c b = m k \frac{a}{d}= \frac{c}{b}=\frac{m}{k} . So for any integers p , q p,q , the solution of a b = c d ab=cd is ( a , b , c , d ) = ( m p , k q , m q , k p ) (a,b,c,d)=(mp,kq,mq,kp) . So a n + b n + c n + d n a^{n}+b^{n}+c^{n}+d^{n} is equal to ( m n + k n ) ( p n + q n ) (m^{n}+k^{n})(p^{n}+q^{n}) , on factorising . Indeed this is composite and each term is greater than 1 1 .

Patrick Corn
Jul 11, 2014

Writing p = a + b + c + b c / a p = a+b+c+bc/a and simplifying, we get p a = ( a + b ) ( a + c ) pa = (a+b)(a+c) . Since p ( a + b ) ( a + c ) p | (a+b)(a+c) , p ( a + b ) p|(a+b) or p ( a + c ) p | (a+c) . But clearly p = a + b + c + d p = a+b+c+d is strictly greater than both of these, so this is impossible.

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