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Question Number 10944 by 314159 last updated on 03/Mar/17
Find all ordered pairs (a,b) so that ((ab)/(a+b)) is an integer. (a and b are integers).
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{ordered}\:\mathrm{pairs}\:\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{so}\:\mathrm{that}\:\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$$$\left(\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{integers}\right). \\ $$
Commented by FilupS last updated on 03/Mar/17
n=((ab)/(a+b)), n, a, b ∈ Z n=((ab)/(a+b)) ⇒ ab≥a+b ∵∀(ab<a+b) ⇒ 0<n<1 ⇒ n∉Z ab≥a+b a(b−1)=b a=(b/(b−1)) ∴ (b/(b−1))∈Z if b=2 ⇒ a=(2/(2−1))=2 ∴(2, 2) is a solution. I am unsure if this is the only solution because this is a bad attempt
$${n}=\frac{{ab}}{{a}+{b}},\:\:\:\:\:{n},\:{a},\:{b}\:\in\:\mathbb{Z} \\ $$$$\: \\ $$$${n}=\frac{{ab}}{{a}+{b}}\:\:\:\:\Rightarrow\:\:\:\:{ab}\geqslant{a}+{b} \\ $$$$\because\forall\left({ab}<{a}+{b}\right)\:\Rightarrow\:\mathrm{0}<{n}<\mathrm{1}\:\Rightarrow\:{n}\notin\mathbb{Z} \\ $$$$\: \\ $$$${ab}\geqslant{a}+{b} \\ $$$${a}\left({b}−\mathrm{1}\right)={b} \\ $$$${a}=\frac{{b}}{{b}−\mathrm{1}} \\ $$$$\therefore\:\frac{{b}}{{b}−\mathrm{1}}\in\mathbb{Z} \\ $$$$\mathrm{if}\:{b}=\mathrm{2}\:\:\:\Rightarrow\:\:\:{a}=\frac{\mathrm{2}}{\mathrm{2}−\mathrm{1}}=\mathrm{2} \\ $$$$\: \\ $$$$\therefore\left(\mathrm{2},\:\mathrm{2}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{unsure}\:\mathrm{if}\:\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{solution} \\ $$$$\mathrm{because}\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{bad}\:\mathrm{attempt} \\ $$
Commented by mrW1 last updated on 04/Mar/17
i think there should be infinite number pairs: 2/2 3/6 4/4 4/12 5/20 6/6 6/12 6/30 7/42 8/8 8/24 8/56 9/18 9/72 10/10 10/15 10/40 11/110 12/12 12/24 12/36 12/60 12/132 13/156 14/14 14/84 14/182 ......
$${i}\:{think}\:{there}\:{should}\:{be}\:{infinite}\:{number}\:{pairs}: \\ $$$$\mathrm{2}/\mathrm{2} \\ $$$$\mathrm{3}/\mathrm{6} \\ $$$$\mathrm{4}/\mathrm{4}\:\:\:\:\mathrm{4}/\mathrm{12} \\ $$$$\mathrm{5}/\mathrm{20} \\ $$$$\mathrm{6}/\mathrm{6}\:\:\:\:\mathrm{6}/\mathrm{12}\:\:\:\:\mathrm{6}/\mathrm{30} \\ $$$$\mathrm{7}/\mathrm{42} \\ $$$$\mathrm{8}/\mathrm{8}\:\:\:\:\mathrm{8}/\mathrm{24}\:\:\:\:\mathrm{8}/\mathrm{56} \\ $$$$\mathrm{9}/\mathrm{18}\:\:\:\:\mathrm{9}/\mathrm{72} \\ $$$$\mathrm{10}/\mathrm{10}\:\:\:\:\mathrm{10}/\mathrm{15}\:\:\:\:\mathrm{10}/\mathrm{40} \\ $$$$\mathrm{11}/\mathrm{110} \\ $$$$\mathrm{12}/\mathrm{12}\:\:\:\:\mathrm{12}/\mathrm{24}\:\:\:\:\mathrm{12}/\mathrm{36}\:\:\:\:\mathrm{12}/\mathrm{60}\:\:\:\mathrm{12}/\mathrm{132} \\ $$$$\mathrm{13}/\mathrm{156} \\ $$$$\mathrm{14}/\mathrm{14}\:\:\:\:\mathrm{14}/\mathrm{84}\:\:\:\:\mathrm{14}/\mathrm{182} \\ $$$$…… \\ $$
Answered by mrW1 last updated on 04/Mar/17
let ((ab)/(a+b))=n let b≥a and b=ka (k≥1) ⇒((aka)/((1+k)a))=n a=((1+k)/k)×n=n+(n/k) with (n/k)=i or k=(n/i) ⇒a=n+i b=ka=(n/i)(n+i)=n(1+(n/i)) with (n/i)=j or n=ji ⇒a=n+i=ji+i=(1+j)i ⇒b=n(1+j)=j(1+j)i i.e. the solution is { ((a=(1+j)i)),((b=j(1+j)i)) :} i,j∈N e.g. i=1, j=1,2,3,4... a/b=2/2, 3/6, 4/12, 5/20, 6/30...... i=2, j=1,2,3,4... a/b=4/4, 6/12, 8/24, 10/40, 12/60...... i=3, j=1,2,3,4... a/b=6/6, 9/18, 12/36, 15/60, 18/90...... ..... We see in the solution that a is the product of 2 numbers and b is the product of 3 numbers and b=a×j. For every a ≥2, if a is a prime number, there is one corresponding value for b, and if a is no prime number, there are two or more corresponding values for b. E.g. a=7 (prime number) which can only be expressed as 7×1 (j=6) ⇒there is only one value for b: b=7×6=42 E.g. a=12 which can be expressed as 12×1 (j=11) 3×4 (j=2) 4×3 (j=3) 2×6 (j=1) 6×2 (j=5) ⇒there are 5 values for b: b=12×1=12 b=12×2=24 b=12×3=36 b=12×5=60 b=12×11=132 the pairs of a and b for a from 2 till 14 see comment above.
$${let}\:\frac{{ab}}{{a}+{b}}={n} \\ $$$${let}\:{b}\geqslant{a}\:{and}\:{b}={ka}\:\left({k}\geqslant\mathrm{1}\right) \\ $$$$\Rightarrow\frac{{aka}}{\left(\mathrm{1}+{k}\right){a}}={n} \\ $$$${a}=\frac{\mathrm{1}+{k}}{{k}}×{n}={n}+\frac{{n}}{{k}} \\ $$$${with}\:\frac{{n}}{{k}}={i}\:{or}\:{k}=\frac{{n}}{{i}} \\ $$$$\Rightarrow{a}={n}+{i} \\ $$$${b}={ka}=\frac{{n}}{{i}}\left({n}+{i}\right)={n}\left(\mathrm{1}+\frac{{n}}{{i}}\right) \\ $$$${with}\:\frac{{n}}{{i}}={j}\:{or}\:{n}={ji} \\ $$$$\Rightarrow{a}={n}+{i}={ji}+{i}=\left(\mathrm{1}+{j}\right){i} \\ $$$$\Rightarrow{b}={n}\left(\mathrm{1}+{j}\right)={j}\left(\mathrm{1}+{j}\right){i} \\ $$$$ \\ $$$${i}.{e}.\:{the}\:{solution}\:{is} \\ $$$$\begin{cases}{{a}=\left(\mathrm{1}+{j}\right){i}}\\{{b}={j}\left(\mathrm{1}+{j}\right){i}}\end{cases} \\ $$$${i},{j}\in\mathbb{N} \\ $$$$ \\ $$$${e}.{g}. \\ $$$${i}=\mathrm{1},\:{j}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}… \\ $$$${a}/{b}=\mathrm{2}/\mathrm{2},\:\mathrm{3}/\mathrm{6},\:\mathrm{4}/\mathrm{12},\:\mathrm{5}/\mathrm{20},\:\mathrm{6}/\mathrm{30}…… \\ $$$$ \\ $$$${i}=\mathrm{2},\:{j}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}… \\ $$$${a}/{b}=\mathrm{4}/\mathrm{4},\:\mathrm{6}/\mathrm{12},\:\mathrm{8}/\mathrm{24},\:\mathrm{10}/\mathrm{40},\:\mathrm{12}/\mathrm{60}…… \\ $$$$ \\ $$$${i}=\mathrm{3},\:{j}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}… \\ $$$${a}/{b}=\mathrm{6}/\mathrm{6},\:\mathrm{9}/\mathrm{18},\:\mathrm{12}/\mathrm{36},\:\mathrm{15}/\mathrm{60},\:\mathrm{18}/\mathrm{90}…… \\ $$$$….. \\ $$$$ \\ $$$${We}\:{see}\:{in}\:{the}\:{solution}\:{that}\:{a}\:{is}\:{the} \\ $$$${product}\:{of}\:\mathrm{2}\:{numbers}\:{and}\:{b}\:{is}\:{the} \\ $$$${product}\:{of}\:\mathrm{3}\:{numbers}\:{and}\:{b}={a}×{j}. \\ $$$$ \\ $$$${For}\:{every}\:{a}\:\geqslant\mathrm{2},\: \\ $$$${if}\:{a}\:{is}\:{a}\:{prime}\:{number},\:{there}\:{is}\:{one} \\ $$$${corresponding}\:{value}\:{for}\:{b},\:{and} \\ $$$${if}\:{a}\:{is}\:{no}\:{prime}\:{number},\:{there}\:{are}\:{two} \\ $$$${or}\:{more}\:{corresponding}\:{values}\:{for}\:{b}. \\ $$$$ \\ $$$${E}.{g}.\:{a}=\mathrm{7}\:\left({prime}\:{number}\right)\:{which}\:{can} \\ $$$${only}\:{be}\:{expressed}\:{as} \\ $$$$\mathrm{7}×\mathrm{1}\:\:\:\left({j}=\mathrm{6}\right) \\ $$$$\Rightarrow{there}\:{is}\:{only}\:{one}\:{value}\:{for}\:{b}: \\ $$$${b}=\mathrm{7}×\mathrm{6}=\mathrm{42} \\ $$$$ \\ $$$${E}.{g}.\:{a}=\mathrm{12}\:{which}\:{can}\:{be}\:{expressed}\:{as} \\ $$$$\mathrm{12}×\mathrm{1}\:\:\left({j}=\mathrm{11}\right) \\ $$$$\mathrm{3}×\mathrm{4}\:\:\:\left({j}=\mathrm{2}\right) \\ $$$$\mathrm{4}×\mathrm{3}\:\:\:\left({j}=\mathrm{3}\right) \\ $$$$\mathrm{2}×\mathrm{6}\:\:\:\left({j}=\mathrm{1}\right) \\ $$$$\mathrm{6}×\mathrm{2}\:\:\:\left({j}=\mathrm{5}\right) \\ $$$$\Rightarrow{there}\:{are}\:\mathrm{5}\:{values}\:{for}\:{b}: \\ $$$${b}=\mathrm{12}×\mathrm{1}=\mathrm{12} \\ $$$${b}=\mathrm{12}×\mathrm{2}=\mathrm{24} \\ $$$${b}=\mathrm{12}×\mathrm{3}=\mathrm{36} \\ $$$${b}=\mathrm{12}×\mathrm{5}=\mathrm{60} \\ $$$${b}=\mathrm{12}×\mathrm{11}=\mathrm{132} \\ $$$$ \\ $$$${the}\:{pairs}\:{of}\:{a}\:{and}\:{b}\:{for}\:{a}\:{from}\:\mathrm{2}\:{till}\:\mathrm{14} \\ $$$${see}\:{comment}\:{above}. \\ $$
Commented by FilupS last updated on 04/Mar/17
I Tried a slimilar subtitution, but couldnt work with it.Awesome job
$$\mathrm{I}\:\mathrm{Tried}\:\mathrm{a}\:\mathrm{slimilar}\:\mathrm{subtitution},\:\mathrm{but}\:\mathrm{couldnt} \\ $$$$\mathrm{work}\:\mathrm{with}\:\mathrm{it}.\mathrm{Awesome}\:\mathrm{job} \\ $$

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