{"id":59204,"date":"2016-04-16T12:44:08","date_gmt":"2016-04-16T15:44:08","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=59204"},"modified":"2023-03-31T20:53:16","modified_gmt":"2023-03-31T23:53:16","slug":"desafio-quatro-equacoes-com-raizes-inteiras","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/desafio-quatro-equacoes-com-raizes-inteiras\/","title":{"rendered":".Desafio: Quatro equa\u00e7\u00f5es com ra\u00edzes inteiras"},"content":{"rendered":"<p style=\"margin:20px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nExistem n\u00fameros inteiros positivos [tex]a, \\, b \\, [\/tex] e [tex] \\, c[\/tex] tais que as ra\u00edzes das equa\u00e7\u00f5es:<\/p>\n<ul>\n<li>[tex]ax^2+bx+c=0[\/tex]<\/li>\n<li>[tex]ax^2+bx-c=0[\/tex]<\/li>\n<li>[tex]ax^2-bx+c=0[\/tex] <\/li>\n<li>[tex]ax^2-bx-c=0[\/tex]<\/li>\n<\/ul>\n<p> sejam n\u00fameros inteiros?<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nObserve, inicialmente, que se [tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o n\u00fameros inteiros, n\u00e3o necessariamente distintos, ent\u00e3o:<\/p>\n<ul>\n<li>[tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]\\left(x-x_1\\right) \\, \\left(x-x_2\\right)=0[\/tex]; ou seja,<br \/>\n[tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-(x_1 + x_2)x+x_1 x_2=0[\/tex];<\/li>\n<li>[tex]-x_1[\/tex] e [tex]-x_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]\\left(x+x_1\\right) \\, \\left(x+x_2\\right)=0[\/tex]; ou seja,<br \/>\n[tex]-x_1[\/tex] e [tex]-x_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]x^2+(x_1 + x_2)x+x_1 x_2=0[\/tex];<\/li>\n<li>[tex]x_1 + x_2 \\, [\/tex] e [tex] \\, x_1 x_2[\/tex] s\u00e3o n\u00fameros inteiros.<\/li>\n<\/ul>\n<p>Dessa forma, se [tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o n\u00fameros inteiros, ent\u00e3o [tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o de coeficientes inteiros [tex]x^2-(x_1 + x_2)x+x_1 x_2=0[\/tex], e, automaticamente, os n\u00fameros inteiros [tex]-x_1[\/tex] e [tex]-x_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]x^2+(x_1 + x_2)x+x_1 x_2=0[\/tex].<br \/>\nA partir dessa conclus\u00e3o, podemos tra\u00e7ar um caminho que nos permitir\u00e1 resolver o problema proposto:<\/p>\n<p style=\"color:#800000;margin:10px 35px 10px 35px;\"><strong>encontrar n\u00fameros inteiros [tex]r_1, \\, r_2, \\, s_1, \\, s_2[\/tex] tais que [tex]r_1, \\, r_2[\/tex] sejam as ra\u00edzes de uma equa\u00e7\u00e3o da forma  [tex]ax^2-bx+c=0,[\/tex] com [tex]a, \\, b, \\, c \\, [\/tex] inteiros positivos e [tex]s_1, \\, s_2[\/tex] sejam as ra\u00edzes da equa\u00e7\u00e3o  [tex]ax^2-bx-c=0[\/tex].<\/strong><\/p>\n<p>Dessa forma, automaticamente, [tex]-r_1, \\, -r_2[\/tex] ser\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]ax^2+bx+c=0[\/tex] e [tex]-s_1, \\, -s_2[\/tex] ser\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]ax^2+bx-c=0[\/tex]. Teremos, ent\u00e3o, as equa\u00e7\u00f5es<br \/>\n[tex]\\qquad ax^2-bx+c=0[\/tex] (equa\u00e7\u00e3o 1);<br \/>\n[tex]\\qquad ax^2-bx-c=0[\/tex] (equa\u00e7\u00e3o 2);<br \/>\n[tex]\\qquad ax^2+bx+c=0[\/tex] (equa\u00e7\u00e3o 3);<br \/>\n[tex]\\qquad ax^2+bx-c=0[\/tex] (equa\u00e7\u00e3o 4);<br \/>\ncom ra\u00edzes inteiras, conforme solicitado no problema.<br \/>\nComo [tex]r_1[\/tex] e [tex]r_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-(r_1 + r_2)x+r_1 r_2=0 \\, [\/tex] e [tex]s_1[\/tex] e [tex]s_2[\/tex] s\u00e3o as ra\u00edzes da equa\u00e7\u00e3o [tex]x^2-(s_1 + s_2)x+s_1 s_2=0[\/tex], devemos encontrar inteiros [tex]r_1, \\, r_2, \\, s_1, \\, s_2[\/tex] tais que [tex]r_1 + r_2=s_1 + s_2[\/tex] e [tex]r_1 r_2=-s_1 s_2[\/tex]. Para que [tex]b= r_1 + r_2[\/tex] e [tex]c=r_1 r_2[\/tex] sejam positivos, basta tomarmos [tex]r_1[\/tex] e [tex]r_2[\/tex] positivos.<br \/>\nPara encontrarmos n\u00fameros inteiros [tex]r_1, \\, r_2, \\, s_1, \\, s_2[\/tex] satisfazendo as condi\u00e7\u00f5es exigidas, faremos o seguinte procedimento:<\/p>\n<p style=\"margin-left:30px; margin-right:30px;\"><strong>(1)<\/strong> Para cada n\u00famero natural n\u00e3o nulo [tex]n[\/tex] faremos, se poss\u00edvel, a sua decomposi\u00e7\u00e3o em <strong>diferentes <\/strong>produtos de dois fatores inteiros positivos,  [tex]n=n_in_j \\, [\/tex]. Para facilitar o passo seguinte, consideraremos [tex]n_i \\le n_j \\, [\/tex].<br \/>\n<strong>(2)<\/strong> Para cada duas decomposi\u00e7\u00f5es distintas, digamos [tex]n=n_t n_k \\, [\/tex] e [tex] \\, n=n_m  n_h[\/tex], observaremos se  [tex]n_k + n_t=n_h-n_m[\/tex] ou [tex]n_h + n_m=n_k-n_t[\/tex].<br \/>\n<span style=\"padding-left:30px;\"> &#8211; se [tex]n_k + n_t=n_h-n_m[\/tex], ent\u00e3o podemos tomar [tex]r_1=n_k, \\, r_2=n_t, \\, s_1=n_h, \\, s_2=-n_m[\/tex];<\/span><br \/>\n<span style=\"padding-left:30px;\"> &#8211; se [tex]n_h + n_m=n_k-n_t[\/tex], ent\u00e3o podemos tomar [tex]r_1=n_h, \\, r_2=n_m, \\, s_1=n_k, \\, s_2=-n_t[\/tex].<\/span><br \/>\n<strong>(3)<\/strong> No momento que conseguirmos uma resposta positiva para o item anterior o problema estar\u00e1 resolvido! Mas observe que o passo <strong>(2)<\/strong> n\u00e3o pode ser aplicado quando [tex]n[\/tex] for um n\u00famero primo, pois nesses casos s\u00f3 existe uma decomposi\u00e7\u00e3o de [tex]n[\/tex] nas condi\u00e7\u00f5es de <strong>(1)<\/strong>: [tex]n= 1n[\/tex].<\/p>\n<p>\u00c0 primeira vista parece complicado; mas n\u00e3o \u00e9. Observe alguns exemplos abaixo.<\/p>\n<p style=\"margin:20px\">\n<p><strong>1)<\/strong><\/p>\n<table style=\"border: 1px solid #616161; width:30%; height:auto;text-align:justify;\">\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]4=n_i n_j \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]4=1\\times 4[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]4=2\\times 2[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j + n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]5[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]4[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j &#8211; n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]3[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]0[\/tex]<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p><strong>Conclus\u00e3o:<\/strong> Para [tex]n=4[\/tex] n\u00e3o temos a situa\u00e7\u00e3o descrita no passo (2).<\/p>\n<p style=\"margin:20px\">\n<p><strong>2)<\/strong><\/p>\n<table style=\"border: 1px solid #616161; width:30%; height:auto;text-align:justify;\">\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]6=n_i n_j \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]6=1\\times 6[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]6=2\\times 3[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j + n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]7[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;color: #f32e51;\">[tex]5[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j &#8211; n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;color: #f32e51;\">[tex]5[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]1[\/tex]<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p><strong>Conclus\u00e3o:<\/strong> Para [tex]n=6[\/tex] temos a situa\u00e7\u00e3o descrita no passo (2). Assim:<\/p>\n<ul>\n<li>[tex]r_1=3; \\, r_2=2; \\, s_1=6; \\, s_2=-1;[\/tex]<\/li>\n<li><strong>equa\u00e7\u00e3o 1:<\/strong> [tex] x^2-5x+6; \\,  \\,  \\,  \\,  \\,  \\,  \\, [\/tex]  <strong>equa\u00e7\u00e3o 2:<\/strong> [tex] x^2-5x-6; \\,  \\,  \\,  \\,  \\,  \\,  \\, [\/tex]  <strong>equa\u00e7\u00e3o 3:<\/strong> [tex] x^2+5x+6; \\,  \\,  \\,  \\,  \\,  \\,  \\, [\/tex]  <strong>equa\u00e7\u00e3o 4:<\/strong> [tex] x^2+5x-6. \\,  \\,  \\,  \\,  \\,  \\,  \\, [\/tex]<\/li>\n<li><strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 1:<\/strong> <span class=\"lat\">2<\/span> e <span class=\"lat\">3<\/span>;<br \/>\n<strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 2:<\/strong> <span class=\"lat\">-1<\/span> e <span class=\"lat\">6<\/span>;<br \/>\n<strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 3:<\/strong> <span class=\"lat\">-2<\/span> e <span class=\"lat\">-3<\/span>;<br \/>\n <strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 4:<\/strong> <span class=\"lat\">1<\/span> e <span class=\"lat\">-6<\/span>. <\/li>\n<\/ul>\n<p style=\"margin:20px\">\n<p><strong>3)<\/strong><\/p>\n<table style=\"border: 1px solid #616161; width:51%; height:auto;text-align:justify;\">\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]18=n_i n_j \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]18=1\\times 18[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]18=2\\times 9[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]18=3\\times 6[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j + n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]19[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]11[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]9[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j &#8211; n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]17[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]7[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]3[\/tex]<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p><strong>Conclus\u00e3o:<\/strong> Para [tex]n=18[\/tex] n\u00e3o temos a situa\u00e7\u00e3o descrita no passo (2).<\/p>\n<p style=\"margin:20px\">\n<p><strong>4)<\/strong><\/p>\n<table style=\"border: 1px solid #616161; width:61%; height:auto;text-align:justify;\">\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]24=n_i n_j \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]24=1\\times 24[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]24=2\\times 12[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]24=3\\times 8[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]24=4\\times 6[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j + n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]25[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]14[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]11[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;color: #f32e51;\">[tex]10[\/tex]<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td style=\"border: 1px solid #616161;\">\u00a0[tex]n_j &#8211; n_i \\, [\/tex]<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]23[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;color: #f32e51;\">[tex]10[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]5[\/tex]<\/p>\n<\/td>\n<td style=\"border: 1px solid #616161;\">\n<p style=\"text-align:center;\">[tex]2[\/tex]<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p><strong>Conclus\u00e3o:<\/strong> Para [tex]n=24[\/tex] temos a situa\u00e7\u00e3o descrita no passo (2). Assim:<\/p>\n<ul>\n<li>[tex]r_1=6; \\, r_2=4; \\, s_1=12; \\, s_2=-2;[\/tex]<\/li>\n<li><strong>equa\u00e7\u00e3o 1:<\/strong> [tex] x^2-10x+24; \\,  \\,  \\,  \\,  \\, [\/tex]  <strong>equa\u00e7\u00e3o 2:<\/strong> [tex] x^2-10x-24; \\,  \\,  \\,  \\,  \\, [\/tex]  <strong>equa\u00e7\u00e3o 3:<\/strong> [tex] x^2+10x+24; \\,  \\,  \\,  \\,  \\, [\/tex]  <strong>equa\u00e7\u00e3o 4:<\/strong> [tex] x^2+10x-24.[\/tex]<\/li>\n<li><strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 1:<\/strong> <span class=\"lat\">4<\/span> e <span class=\"lat\">6<\/span>;<br \/>\n<strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 2:<\/strong> <span class=\"lat\">-2<\/span> e <span class=\"lat\">12<\/span>;<br \/>\n<strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 3:<\/strong> <span class=\"lat\">-4<\/span> e <span class=\"lat\">-6<\/span>;<br \/>\n <strong>solu\u00e7\u00f5es da equa\u00e7\u00e3o 4:<\/strong> <span class=\"lat\">2<\/span> e <span class=\"lat\">-12<\/span>. <\/li>\n<\/ul>\n<p style=\"margin:20px\">\n<p>Portanto, existem n\u00fameros inteiros positivos que satisfazem as condi\u00e7\u00f5es do problema. Por exemplo:<br \/>\n[tex] \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\, \\boxed{a=1, \\, b=5, \\, c=6} \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\, [\/tex]e [tex] \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\,  \\, \\boxed{a=1, \\, b=10, \\, c=24}[\/tex].<br \/>\nAl\u00e9m desse dois exemplos, existem outros. <\/p>\n<p class=\"ti3 c3\"  style=\"line-height:30px;margin:10px\">Que tal procurar mais exemplos considerando [tex]n[\/tex] com dois d\u00edgitos?<\/p>\n<p>Para quem aceitou o convite, <a href=\"http:\/\/clubes.obmep.org.br\/blog\/texto_012-contagem-de-divisores-um-segundo-estudo\/\" target=\"_blank\"><strong>este texto<\/strong><\/a> pode ser \u00fatil!<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) Existem n\u00fameros inteiros positivos [tex]a, \\, b \\, [\/tex] e [tex] \\, c[\/tex] tais que as ra\u00edzes das equa\u00e7\u00f5es: [tex]ax^2+bx+c=0[\/tex] [tex]ax^2+bx-c=0[\/tex] [tex]ax^2-bx+c=0[\/tex] [tex]ax^2-bx-c=0[\/tex] sejam n\u00fameros inteiros? Solu\u00e7\u00e3o Observe, inicialmente, que se [tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o n\u00fameros inteiros, n\u00e3o necessariamente distintos, ent\u00e3o: [tex]x_1[\/tex] e [tex]x_2[\/tex] s\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/desafio-quatro-equacoes-com-raizes-inteiras\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-59204","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/59204","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=59204"}],"version-history":[{"count":10,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/59204\/revisions"}],"predecessor-version":[{"id":185987,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/59204\/revisions\/185987"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=59204"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}