Tìm n thuộc N để (n^2-3)^2 +16 là 1 số chính phương

1 câu trả lời

(n23)2+16=k2(kZ)k2(n23)2=16(kn2+3)(k+n23)=16Dok,nZkn2+3U(16)={±1;±2;±4;±8;±16}TH1:{kn2+3=1k+n23=16{k=8n2=10(Loai)TH2:{kn2+3=1k+n23=16{k=172n2=92(Loai)TH3:{kn2+3=2k+n23=8{k=5n2=6(Loai)TH4:{kn2+3=2k+n23=8{k=5n2=0n=0(tm)TH5:{kn2+3=4k+n23=4{k=0n2=1(Loai)TH6:{kn2+3=4k+n23=4{k=0n2=7(Loai)TH7:{kn2+3=8k+n23=2{k=5n2=0n=0(tm)TH8:{kn2+3=8k+n23=2{k=5n2=6(Loai)TH9:{kn2+3=16k+n23=1{k=172n2=92(Loai)TH10:{kn2+3=16k+n23=1{k=172n2=212(Loai)Vayn=0