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Question Number 13395 by Tinkutara last updated on 19/May/17
Find all positive integers n for which n^2 + 96 is a perfect square.
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:{n}\:\mathrm{for}\:\mathrm{which} \\ $$$${n}^{\mathrm{2}} \:+\:\mathrm{96}\:\mathrm{is}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}. \\ $$
Answered by ajfour last updated on 19/May/17
(n+m)^2 =n^2 +96 2nm+m^2 −96=0 n=((96−m^2 )/(2m)) ⇒ m should be even and m <10 for m=2, n=23 m=4, n=10 m=6, n=5 m=8, n=2 n=2, 5, 10, 23 .
$$\left({n}+{m}\right)^{\mathrm{2}} ={n}^{\mathrm{2}} +\mathrm{96} \\ $$$$\mathrm{2}{nm}+{m}^{\mathrm{2}} −\mathrm{96}=\mathrm{0} \\ $$$${n}=\frac{\mathrm{96}−{m}^{\mathrm{2}} }{\mathrm{2}{m}}\:\: \\ $$$$\Rightarrow\:{m}\:{should}\:{be}\:{even}\:{and}\:\:{m}\:<\mathrm{10} \\ $$$${for}\:{m}=\mathrm{2},\:{n}=\mathrm{23} \\ $$$$\:\:\:\:\:\:\:\:{m}=\mathrm{4},\:{n}=\mathrm{10} \\ $$$$\:\:\:\:\:\:\:\:{m}=\mathrm{6},\:{n}=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:{m}=\mathrm{8},\:{n}=\mathrm{2} \\ $$$${n}=\mathrm{2},\:\mathrm{5},\:\mathrm{10},\:\mathrm{23}\:. \\ $$
Commented by Tinkutara last updated on 19/May/17
Why n^2 + 96 = (n + m)^2 ?
$$\mathrm{Why}\:{n}^{\mathrm{2}} \:+\:\mathrm{96}\:=\:\left({n}\:+\:{m}\right)^{\mathrm{2}} ? \\ $$
Commented by ajfour last updated on 19/May/17
if n^2 +96 is a perfect square it is greater than n^2 and for m∈N it can be written (m+n)^2 .
$${if}\:{n}^{\mathrm{2}} +\mathrm{96}\:{is}\:{a}\:{perfect}\:{square}\:{it}\:{is} \\ $$$${greater}\:{than}\:{n}^{\mathrm{2}} \:{and}\:{for}\:{m}\in\mathbb{N} \\ $$$${it}\:{can}\:{be}\:{written}\:\left({m}+{n}\right)^{\mathrm{2}} \:. \\ $$
Answered by mrW1 last updated on 19/May/17
n^2 +96=m^2 m^2 −n^2 =96 (m−n)(m+n)=96=i×j ⇒m−n=i ⇒m+n=j ⇒m=((i+j)/2) ⇒n=((j−i)/2) for m and n to be integer, i and j must be both odd or both even. 96=1×96=2×48=3×32=4×24=6×16=8×12 (only red numbers fulfill the condition) for i×j=2×48 ⇒m=((2+48)/2)=25 and n=((48−2)/2)=23 for i×j=4×24 ⇒m=((4+24)/2)=14 and n=((24−4)/2)=10 for i×j=6×16 ⇒m=((6+16)/2)=11 and n=((16−6)/2)=5 for i×j=8×12 ⇒m=((8+12)/2)=10 and n=((12−8)/2)=2 ⇒n=2, 5, 10, 23
$${n}^{\mathrm{2}} +\mathrm{96}={m}^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −{n}^{\mathrm{2}} =\mathrm{96} \\ $$$$\left({m}−{n}\right)\left({m}+{n}\right)=\mathrm{96}={i}×{j} \\ $$$$\Rightarrow{m}−{n}={i} \\ $$$$\Rightarrow{m}+{n}={j} \\ $$$$\Rightarrow{m}=\frac{{i}+{j}}{\mathrm{2}} \\ $$$$\Rightarrow{n}=\frac{{j}−{i}}{\mathrm{2}} \\ $$$${for}\:{m}\:{and}\:{n}\:{to}\:{be}\:{integer},\:{i}\:{and}\:{j}\:{must} \\ $$$${be}\:{both}\:{odd}\:{or}\:{both}\:{even}. \\ $$$$ \\ $$$$\mathrm{96}=\mathrm{1}×\mathrm{96}=\mathrm{2}×\mathrm{48}=\mathrm{3}×\mathrm{32}=\mathrm{4}×\mathrm{24}=\mathrm{6}×\mathrm{16}=\mathrm{8}×\mathrm{12} \\ $$$$\left({only}\:{red}\:{numbers}\:{fulfill}\:{the}\:{condition}\right) \\ $$$$ \\ $$$${for}\:{i}×{j}=\mathrm{2}×\mathrm{48} \\ $$$$\Rightarrow{m}=\frac{\mathrm{2}+\mathrm{48}}{\mathrm{2}}=\mathrm{25}\:{and}\:{n}=\frac{\mathrm{48}−\mathrm{2}}{\mathrm{2}}=\mathrm{23} \\ $$$$ \\ $$$${for}\:{i}×{j}=\mathrm{4}×\mathrm{24} \\ $$$$\Rightarrow{m}=\frac{\mathrm{4}+\mathrm{24}}{\mathrm{2}}=\mathrm{14}\:{and}\:{n}=\frac{\mathrm{24}−\mathrm{4}}{\mathrm{2}}=\mathrm{10} \\ $$$$ \\ $$$${for}\:{i}×{j}=\mathrm{6}×\mathrm{16} \\ $$$$\Rightarrow{m}=\frac{\mathrm{6}+\mathrm{16}}{\mathrm{2}}=\mathrm{11}\:{and}\:{n}=\frac{\mathrm{16}−\mathrm{6}}{\mathrm{2}}=\mathrm{5} \\ $$$$ \\ $$$${for}\:{i}×{j}=\mathrm{8}×\mathrm{12} \\ $$$$\Rightarrow{m}=\frac{\mathrm{8}+\mathrm{12}}{\mathrm{2}}=\mathrm{10}\:{and}\:{n}=\frac{\mathrm{12}−\mathrm{8}}{\mathrm{2}}=\mathrm{2} \\ $$$$ \\ $$$$\Rightarrow{n}=\mathrm{2},\:\mathrm{5},\:\mathrm{10},\:\mathrm{23} \\ $$
Commented by Nayon last updated on 20/May/17
i understand it clerly mr.w1
$${i}\:{understand}\:{it}\:{clerly}\:{mr}.{w}\mathrm{1} \\ $$

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