Duke Mth Meet 01-14 Power Round Qudrtic Residues nd Prime Numers For integers nd, we write to indicte tht evenly divides, nd to indicte tht does not divide For exmle, 4 nd 4 Let e rime numer An integer is clled qudrtic residue modulo if there exists n integer x with x For exmle, if we tke, then 0, 1, nd 4 re qudrtic residues modulo, s 0 0 1 1 4 1 1 oint Exlin why for every integer x, there must e n integer k such tht x is equl to one of k, k + 1, k +, k +, or k + 4 Solution Using division with reminder, we cn write x k + r, where 0 r 4 Thus we hve x in the desired form 1 oint Exlin why every integer of the form k, k + 1, or k + 4 is qudrtic residue modulo Solution For k we my tke x 0 in the definition of qudrtic residue; we then hve x k s desired For k +1 we my tke x 1, s we otin 1 k +1 k Similrly for k + 4 we my tke x c oints Using rt, show tht nd re not qudrtic residues modulo Exlin why every numer of the form k + or k + is not qudrtic residue modulo Solution We show tht is not qudrtic residue modulo y contrdiction Suose there exists x such tht x Write x k + r, so tht x k +10kr+r k +kr+r Then we must hve k +kr+r, so tht r But we only hve ossiilities for r, none of which work: does not divide -, -1,, 7, or 14 Hence cnnot e qudrtic residue modulo An nlogous rgument shows tht is not qudrtic residue modulo either As x iff x +k, relcing y +k or y +k in the ove rguments doesn t chnge their vlidity Hence no numer of the form k + or k + is qudrtic residue modulo Given nd s ove, we write 1 if is qudrtic residue modulo nd ; 0 if if is not qudrtic residue modulo This nottion is commonly clled the Legendre symol Do not confuse this with the frction /! 1 1 oint Comute nd 7 1 Yeh, this nottion isn t the est Unfortuntely, it s trditionl 1
Solution By 1c, we know tht is not qudrtic residue modulo Hence We hve 7, so is qudrtic residue modulo 7 Hence 7 1 1 oint Exlin why 1 for ll rimes nd integers with Solution Tking x in the definition of qudrtic residue, we hve 0 for ll rimes nd integers Hence is lwys qudrtic residue modulo ; if we further ssume tht then 1 c oints Show tht if, then Solution Suose tht 1 Then is qudrtic residue modulo, so there exists x with x Hence x + x, so is qudrtic residue modulo Furthermore, if nd, then, so 1 If 0, then Hence +, so 0 If, then there does not exist x with x If there existed x with x, then we would hve x + x, contrdiction Hence is not qudrtic residue modulo, nd so oints Suose tht > Exlin why exctly + 1/ of the numers {0, 1,,, } re qudrtic residues modulo Hint: if is qudrtic residue, fctor the olynomil x Solution Consider the irs 0, 0, 1, 1,, x, x,, 1, 1, where x is the unique numer etween 0 nd 1 such tht x x For exmle, s 1 1, we hve 1 1 Then the numer of qudrtic residues mong {0, 1,,, 1} is clerly the numer of distinct second elements mong ll these irs Clerly x x, s x x + x Hence 1 nd 1 hve the sme second element, nd similrly for, nd so on There re 1/ of these irs, nd ll of them hve nonzero second element, s if x then x Now we clim tht no other irs hve equl second element For if x y hve equl second elements x, y, then 0 x y Hence x y x yx + y, nd thus either x y or x + y Note tht s x, y {0,, 1} we hve 0 < x y < nd thus x y Hence x + y, nd thus y x So no other irs hve equl second elements Throwing in the second element of zero from the ir 0, 0 does not give us ny more collisions, s noted ove, nd hence there re 1/ + 1 + 1/ qudrtic residues mong {0, 1,,, 1} 4 4 oints Using the result of question, show tht for ny rime numer there must exist ositive integers, with + + 1 Solution If the result is cler: tke even nd odd Now if >, then exctly + 1/ elements of the set {0, 1,,, 1} re qudrtic residues The m x 1 x ms this set to itself ijectively, nd hence y the igeonhole rincile there exists c such tht c nd c re qudrtic residues Thus there exist integers, such tht c nd 1 c
Hence we hve c + + 1 + c + + 1, nd hence + + 1 We my clerly tke, to e ositive A celerted theorem of Euler gives somewht convenient wy to clculte Legendre symols: Euler s Criterion Let > e rime, nd let e n integer Then / To see how to use this to comute Legendre symols, let s clculte We know tht 1 must e divisile y As must e 1 or -1, it follows tht Hence is not qudrtic residue modulo oints Show tht 1 if or is of the form 4k + 1 nd if is of the form 4k + Solution Clerly 1; now suose > Then we hve / As > the only wy tht cn divide the difference of nd / is if they re equl to ech other Hence we hve / Thus if 4k +1 then k 1, nd if 4k + then k+1 6 oints Show tht Solution The sttement is ovious for - there re only 4 cses to check nd they re ll immeditely cler Now suose > Then we hve /, nd hence / Similrly, /, nd hence / / / Hence we hve [ / We lso hve [ ] [ + / /, nd thus ] / ] / / [ ] / / As > nd s the two terms in the rightmost exression re equl to ±1, they must e equl Hence s desired 7 6 oints Let e rime of the form 4k + Using the ove results, show tht if there exist integers, with +, then nd Hint: how re nd relted?
4 Solution Suose tht, ; tht is,, 0 As +, we know tht is qudrtic residue modulo, nd so 1 By ove we hve We hve 1, nd thus 1, contrdicting the hyothesis tht 4k + Hence it must e the cse tht, The second fmous theorem concerning the Legendre symol is generlly credited to Guss, nd is known s the lw of qudrtic recirocity: Qudrtic Recirocity Let q e odd rime numers Then q q 4 q This theorem cn e extended to the cse q nd odd, in which cse it gives 8 8 6 oints Clculte, with exlntion, 4 017 Solution We hve 4 7 017 017 017 017 We hve 16 017 1, nd hence 16 017 1 so tht 017 1/8 Hence 017 1 y the q cse ove Now we turn to the 017 term As 8 017 1, y qudrtic recirocity we hve 017 017 1 As 017 1 1 it follows tht 017 1 Finlly we clculte 7 017 We hve similrly to the cse tht 7 017 017 7 1 As 016 100 84, we hve 017 7 1 7 1 Hence 7 017 1, nd thus 4 017 1 9 7 oints Show tht if is rime nd n is n integer with n + n + 1, then either or 6k + 1 for some ositive integer k Hint: multily y 4 Solution Note tht n + n + 1 is odd for ll n, nd so Now suose > Tking the hint, we hve 4n + 4n + 4 n + 1 + Hence is qudrtic residue modulo Hence we hve 1 / By qudrtic recirocity we hve /4 ; s ±1 we hve thus / Hence 1 /, nd thus is qudrtic residue modulo Hence j + 1 As must e odd, we must hve j k even, nd thus 6k + 1
10 8 oints Let k e n integer, nd suose tht is n odd rime with k + 1 Show tht the tens digit of must e even Hint: wht must e? Solution Note tht if then the hyothesis is trivilly stisfied Hence suose > Using the sme trick s in the lst rolem, we hve k +, nd thus 1 We hve, nd thus s / we hve y qudrtic recirocity tht 1 / / Hence we hve two cses: first, tht 4 nd 1, nd second, tht ut 4 1 nd Note tht 1 4 1 nd Hence in the first cse we hve either 0k +1 or 0k +9, nd in the second cse we hve either 0k + or 0k + 7 Thus the tens digit of must e odd, s desired