Is there an easy way to find examples of such triples? Why yes! Just look at an ordinary multiplication table to find them!

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1 PUSHING PYTHAGORAS 009 Jmes Tnton A triple of integers ( bc,, ) is clled Pythgoren triple if exmple, some clssic triples re ( 3,4,5 ), ( 5,1,13 ), ( ) fond of ( 0,1,9 ) nd ( 119,10,169 ). + b = c. For 7,4,5. I m personlly Is there n esy wy to find exmples of such triples? Why yes! Just look t n ordinry multipliction tble to find them! Choose ny two numbers on the min digonl (these re squre numbers) nd the two identicl numbers to mke squre of chosen figures. Sum the two squre numbers, tke the difference of the two squre numbers, nd sum the two identicl numbers. You now hve Pythgoren triple! e.g. = 5 4= 1 b= = = 9 c= 5+ 4= 9 As nother exmple, choose 36 nd 1 to obtin: = 36 1= 35 b= 6+ 6= = 37 c= 36+ 1= 37 Question 1: Which two squre numbers give the triple (3,4,5)? Which give the triples (5, 1, 13) nd (7, 4, 5)? Question : Why does this trick work? Tough Chllenge: This method fils to yield (9,1,15) but if the common fctor of three is removed we do obtin (3,4,5). Show tht every Pythgoren triple with no shred fctors between the three terms does indeed pper vi this method. 009 Jmes Tnton

2 PYTHAGORAS MEETS THE THIRD DIMENSION A clssic problem in geometry is to work out the length of the longest digonl in rectngulr box. For exmple, wht is the distnce between points A nd B in this box? One ppliction of the Pythgoren theorem shows tht the length of the digonl on the bse of the figure is 5 units long. A second ppliction of the theorem using the shded tringle yields the length we seek s 13 units. In generl, precisely this rgument shows tht the length of the longest digonl d in n b c rectngulr box stisfies: d = + b + c This is just one wy to think of the Pythgoren theorem extended to the third dimension. Consider the tringle sitting inside rectngulr box s shown. Wht is its re? We cn compute this with reltive ese. (The lgebr is td messy but it is worth plowing through.). Strt by removing the three right tringles bout the top front corner of the box to give clerer view. Drw in n ltitude for the tringle nd lbel the lengths x, y, nd h s shown: Do you see tht x + h = b + c nd y+ x= + c? Do you lso see tht y + h = + b? Let s ply with this third eqution using the first two for support. 009 Jmes Tnton

3 We hve: So: yielding: ( ) + c x + h = + b + c x + c + x + h = + b x + b = x + h + c b = c Thus: c x= + c Now solve for h : 4 c h = b + c x = b + c + c b + bc + c = + c (Yick!) 1 Now we re ll set to work out the re of the tringle: A bse times height. ) To void squre roots let s compute insted = + c h. ( Hlf A : 1 1 A = ( + c ) h = ( b + bc + c ) = b + bc + c Notice tht ech term squred is the re of one of the right tringles on the side of the rectngulr box. So wht hve we got? A tringle of re A sits cross three mutully perpendiculr right tringles. If these tringles hve res B, C, nd D, then: A = B + C + D. Exercise: A tringle crosses the x-, y- nd z-xes t positions 3, 5 nd 8. Wht is its re? This lterntive 3-D version of Pythgors s theorem is not prticulrly well known! 009 Jmes Tnton

4 RESEARCH CORNER: The following box hs the property tht ech side-length is n integer nd ech digonl cross fce is n integer. Chllenge1: Find nother exmple of such box. Chllenge : Unfortuntely, the length of the longest digonl inside this box, 735, is not n integer. No one on this plnet currently knows n exmple of box with ll side lengths nd ll digonls integers. For world fme, cn you find one? HARD STUFF: ****** WHY DOES THE MULTIPLICATION TABLE TRICK WORK? The clim is tht every Pythgoren triple ( bc,, ) with + b = c nd, b nd c shring no common fctor cn be written in the form: = m n b= mn c= m + n For some pir of integers m ndn. e.g. (3,4,5) comes from choosing m = nd n = 1. (5,1,13) from m = 3 nd n =. (7, 4, 5) from m = 4 nd n = 3. It is esy to see tht if, b nd c re of this form, then we hve Pythgoren triple: 009 Jmes Tnton

5 + b = ( m n ) + ( mn) = m + n mn + 4mn 4 4 = m + n + mn 4 4 = ( m + n ) = c Proving tht every triple (,b,c) with no common fctors hs entries of this form is tricky. Here s proof: THEOREM: Suppose common fctors. Then Proof: + b = c with ( bc,, ) triple of integers with no = m n b= mn c= m + n 1. The numbers, b, nd c cnnot ll be even. (Otherwise they hve common fctor of two.) So t lest one of the numbers is odd.. If nd b re both even, then c = + b is even, mking c even. We cnnot hve this. So one of or b (or both) is odd. Without loss of generlity, let s sy tht is n odd integer. 3. If b is odd, then b k r ( k r k r) + = ( + 1) + ( + 1) = for some numbers k nd r. This mens tht c is squre number tht is two more thn multiple of 4. This is impossible! Reson: Either c is multiple of four, it is one more thn multiple of four, two more thn multiple of four or three more thn multiple of four. In ll these cses, we never get tht c is two more thn multiple of four! 009 Jmes Tnton

6 ( q) = 4q (q+ 1) = 4( q + q) + 1 (q+ ) = 4( q+ 1) q q q ( + 3) = 4( ) + 1 So we must hve odd, b even mking c odd. 4. Now: + b = c which mens: ( )( ) b = c = c c+ b c c+ But b is even nd so is divisible by two. So = eqution involving integers. (Recll tht nd c re odd.) is vlid Now comes the relly tricky prt! c c+ 5. If nd re both multiples of common number k then their sum, which is c nd their difference, which is, would lso be multiples of k. And since b = c, b would be lso be multiple of k nd AFTER SOME TRICKY THOUGHT ABOUT HOW FACTORS OF NUMBERS WORK, this mens tht b itself would hve to be multiple of k. This is impossible since, b nd c hve no common fctors except k = 1. So: c nd c+ hve no common fctors. 6. SOME MORE TRICKY THOUGHT ABOUT HOW FACTORS WORK gives Since c nd c+ hve no fctor in common yet their product is 009 Jmes Tnton

7 squre number (nmely c+ re themselves squre numbers: b ), this cn only hppen if ech of c nd c = n c+ = m Adding nd subtrcting gives c = m + n = m n Solving for b gives: b c mn = =. We re done! COMMENT: The tricky number theory used here is bsed on the following principles: 1. Every number breks down into product of primes. Thus when consider common fctor k of set of numbers, it suffices to ssume we re considering fctor tht is prime number.. Primes p hs the property tht if product MN is multiple of p, then either M is multiple of p or N is (or both). Thus, ifr = MN, with M nd N shring no common fctors, then ny prime fctor p of R either goes into M or N (but not both). And since p goes into R, then this mens tht p ctully goes into either M or N. Also, ll the prime fctors of M nd of N go into R nd hence R. 009 Jmes Tnton

8 All in ll, this mens tht the prime fctors of M must come squred mking M squre number, nd the prime fctors of N come squres mking N squre number if we hope R = MN to be true. FINAL COMMENT: All of this tricky number theory is explined with greter detil nd greter clrity in Volume of THINKING MATHEMATICS! vilble for purchse from Jmes Tnton 009 Jmes Tnton