Amicable pairs of numbers and their product

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Suppose that $(a,b)$ ($a<b$) is an amicable pair and $(c,d)$ ($c<d$) another pair such that $$ab=cd$$




Is it possible that (c,d) is an amicable pair as well ? Or is it easy to show that it is impossible ?




In other words : Can distinct amicable pairs have the same product ?




number-theory elementary-number-theory




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  • I dont think that it should say c/d it should say c,d
    – argamon
    Jul 18 at 21:40










  • Upto $10^8$, the pairs give distinct products.
    – Peter
    Jul 21 at 19:16










  • Thanks Peter how are you
    – argamon
    Jul 21 at 19:19










  • @Peter any tips on how to prove it?
    – argamon
    Jul 21 at 19:37










  • I just used brute force, so no.
    – Peter
    Jul 22 at 8:04














up vote
1
down vote

favorite












Suppose that $(a,b)$ ($a<b$) is an amicable pair and $(c,d)$ ($c<d$) another pair such that $$ab=cd$$




Is it possible that (c,d) is an amicable pair as well ? Or is it easy to show that it is impossible ?




In other words : Can distinct amicable pairs have the same product ?




number-theory elementary-number-theory




share|cite|improve this question





















  • I dont think that it should say c/d it should say c,d
    – argamon
    Jul 18 at 21:40










  • Upto $10^8$, the pairs give distinct products.
    – Peter
    Jul 21 at 19:16










  • Thanks Peter how are you
    – argamon
    Jul 21 at 19:19










  • @Peter any tips on how to prove it?
    – argamon
    Jul 21 at 19:37










  • I just used brute force, so no.
    – Peter
    Jul 22 at 8:04












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Suppose that $(a,b)$ ($a<b$) is an amicable pair and $(c,d)$ ($c<d$) another pair such that $$ab=cd$$




Is it possible that (c,d) is an amicable pair as well ? Or is it easy to show that it is impossible ?




In other words : Can distinct amicable pairs have the same product ?




number-theory elementary-number-theory




share|cite|improve this question













Suppose that $(a,b)$ ($a<b$) is an amicable pair and $(c,d)$ ($c<d$) another pair such that $$ab=cd$$




Is it possible that (c,d) is an amicable pair as well ? Or is it easy to show that it is impossible ?




In other words : Can distinct amicable pairs have the same product ?





number-theory elementary-number-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 18 at 21:47









Anonymous

4,8033940




4,8033940









asked Jul 18 at 20:53









argamon

625




625











  • I dont think that it should say c/d it should say c,d
    – argamon
    Jul 18 at 21:40










  • Upto $10^8$, the pairs give distinct products.
    – Peter
    Jul 21 at 19:16










  • Thanks Peter how are you
    – argamon
    Jul 21 at 19:19










  • @Peter any tips on how to prove it?
    – argamon
    Jul 21 at 19:37










  • I just used brute force, so no.
    – Peter
    Jul 22 at 8:04
















  • I dont think that it should say c/d it should say c,d
    – argamon
    Jul 18 at 21:40










  • Upto $10^8$, the pairs give distinct products.
    – Peter
    Jul 21 at 19:16










  • Thanks Peter how are you
    – argamon
    Jul 21 at 19:19










  • @Peter any tips on how to prove it?
    – argamon
    Jul 21 at 19:37










  • I just used brute force, so no.
    – Peter
    Jul 22 at 8:04















I dont think that it should say c/d it should say c,d
– argamon
Jul 18 at 21:40




I dont think that it should say c/d it should say c,d
– argamon
Jul 18 at 21:40












Upto $10^8$, the pairs give distinct products.
– Peter
Jul 21 at 19:16




Upto $10^8$, the pairs give distinct products.
– Peter
Jul 21 at 19:16












Thanks Peter how are you
– argamon
Jul 21 at 19:19




Thanks Peter how are you
– argamon
Jul 21 at 19:19












@Peter any tips on how to prove it?
– argamon
Jul 21 at 19:37




@Peter any tips on how to prove it?
– argamon
Jul 21 at 19:37












I just used brute force, so no.
– Peter
Jul 22 at 8:04




I just used brute force, so no.
– Peter
Jul 22 at 8:04















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