Tìm n thuộc N để (n^2-3)^2 +16 là 1 số chính phương
1 câu trả lời
(n2−3)2+16=k2(k∈Z)⇔k2−(n2−3)2=16⇔(k−n2+3)(k+n2−3)=16Dok,n∈Z⇒k−n2+3∈U(16)={±1;±2;±4;±8;±16}TH1:{k−n2+3=1k+n2−3=16⇔{k=8n2=10(Loai)TH2:{k−n2+3=−1k+n2−3=−16⇔{k=−172n2=−92(Loai)TH3:{k−n2+3=2k+n2−3=8⇔{k=5n2=6(Loai)TH4:{k−n2+3=−2k+n2−3=−8⇔{k=−5n2=0⇔n=0(tm)TH5:{k−n2+3=4k+n2−3=−4⇔{k=0n2=−1(Loai)TH6:{k−n2+3=−4k+n2−3=4⇔{k=0n2=7(Loai)TH7:{k−n2+3=8k+n2−3=2⇔{k=5n2=0⇔n=0(tm)TH8:{k−n2+3=−8k+n2−3=−2⇔{k=−5n2=6(Loai)TH9:{k−n2+3=16k+n2−3=1⇔{k=172n2=−92(Loai)TH10:{k−n2+3=−16k+n2−3=−1⇔{k=−172n2=212(Loai)Vayn=0