Katedra matematiky - Katolícka univerzita v Ružomberku

64 Jan Kopka, Leonard Frobisher, George Feissner3 = 1·3 = 0+1+25 = 1·5 = −1+0+1+2+37 = 1·7 = −2+(−1)+0+1+2+3+410 = 2·5 = 0+1+2+3+414 = 2·7 = −1+0+1+2+3+4+5.However, this last experiment has shown us exactly what we need to do toshow that Conjecture 3a is true. We present it as a proof.Proof of conjecture 3a): Let positive integer number n have an odd factor2k +1. Then n can be written in the form (1). Now we know, that n canbe written also in the formn = (g−k)+···+(g−2)+(g−1)+g+(g+1)+(g+2)+···+(g+k) (2)where g is also factor of n.How can we get only two or more positive numbers on the right hand sideof formula (2)? If there is zero, we delete it. All we have to do is counteractthe negative terms by the corresponding positive terms. There are morepositive terms than negative ones because the total sum is positive. Afterthis procedure we get the even number of consecutive positive terms.So after this proof we rename conjecture 3a as a theorem:Theorem 1: If a positive integer n has an odd factor other than 1, then itcan be written as the sum of consecutive positive integers.Indeed, our investigation has given us more than what the theorem states;not only do we know that it can be done, but we know how to do it!Now we have enough experiences to believe that the opposite theorem totheorem 1 also holds.Theorem 2: If a positive integer n can be written as the sum of consecutivepositive integers, then it must have an odd factor other than 1.Proof: We writenasn = g 1 +g 2 +···+g r whereg 1 ,g 2 ,...g r are consecutivepositive integers. We consider two cases:(a) r is an odd number(b) r is an even number.Case (a): We can write n in the form shown in Formula 1, and then g willbe the number in the middle of the sum. The fact that r is odd guaranteesthere is a middle term. Then, exactly as in our experimentation, we canthen write n in the form n = g? (2k + 1) so n has an odd factor greaterthan 1, namely 2k +1.