Simple theorem on the integral roots of special class of prime degree polynomial equations

Abstract

Even in case of a simple polynomial x3 + l5xb + 28 = 0 , where (3, b) = 1 , it may be extremely difficult to discard the integral solutions without knowing the number b exactly .In this case, one can make use of the method of Tartaglia and Cardan [Archbold J.W.1961] and its solutions can be written as u + v,Ul:o + vw2 ,uw2 + vw, where u3, v3 are the roots of the equation x2 + 28x -125b3 = 0, and {J) is the cube root I . . [-28 ± .J282 + 500b3 J 3 of umty .Also, u or v can be wntten as 2 and this expression is obviously zero only when b = 0. Therefore if b :f. 0 , it is very difficult to determine that I k [-28±.J282+500b3J3. . = Th h 'lib I' d · h 2 IS an mteger or not . e t eorem w1 e exp rune m t e . following , is Capable of discarding all integral solutions of this equation using only one condition (3, b) = 1 . The theorem in its naive form discards all integral solutions of the polynomial . xP + pbx-cP = 0, where p is a prime and (p,b) = (p,c) = 1 Theorem xP + pbx-cP = 0 has no integral solutions if (p,b) = (p,c) = 1 , where b,c are any integers and p is any prime . Proof Proof of the theorem is based on the following lemma Lemma If (a,p) = (b,p) = 1, and ifs= aP-bP is divisible by p , then p2 dividess. This is true even when s = a P + bP and p is odd. Proof of the Lemma s=aP-a-(bP-b)+a-b and since s is divisible by p and aP-a ,bP-b are divisible by p due to Fermat's little theorem, it follows that a-b is divisible by p. aP -bP= (a-b)[(ap-l -bP-1)+b(aP-2 -bP-2)+ .. ·+bP-3(a-b)+ pbP-1)] (1) From ( 1 ), it follows that s is divisible by p2 • Proof of the lemma for a P +bP is almost the above. It is well known that the equation xP + pbx-cP = 0 (2) has either integral or irrational roots. If this equation has an integral root l, let x = l and (p,l) = 1. Then , lP -cP+ pbl= 0 . From the Lemma, it follows that p2 I CF - cP) .Therefore pI b , and this is a contradiction. Therefore equation has no integral roots which are not divisible by p .If it has an integral solution which is divisible by p , then let x=pf3k,(p, k)=l. Then we have, (p/3 k)P + pbp/3 k- cP = 0 ,and hence pI c ,which is again a contradiction since(p, c) = 1 which completes the proof. As an special case of the theorem , consider the equation x3 + 15xb + 28 = 0 (3) which can be written as x3 +1+15xb+33 =0 (4) and it is clear that this equation has no integral root l = O(mod 3) since 1 is not divisible by 3. If this equation has an integral root k which is not divisible by 3 , then k3 + 1 + 15kb + 32 = 0 from which it follows that 31 b due to the Lemma(in case of negative c ) and is a contradiction . Therefore the equation has no integral roots. In case of p = 2 , it follows from the theorem that the equation x2 - 2 bx- q2 =0 where (2, q) = 1 = (2, b) 'has no integral roots. Again from the theorem it follows that x P - pcx - p f3p a P - bP = 0 , where (p, c) = 1 = ( b, p) and p is a prime, has no integral solutions .. In particular here, p f3p a 3, bP are two components o( F ermat triples. It is easy to deduce that this equation has no integral roots. This theorem may hold for some other useful forms of polynomial equations.