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Find all pairs of positive integers $a,b$ such that $$ab=160+ 90\cdot \gcd(a,b)$$ how do we approach this type of problem in number theory or what is the best way to solve this ?Here gcd is greatest common divisor .

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Put $d=\gcd(a,b),a=md,b=nd$ so we have $mnd^2-90d-160=0$. Hence $d$ divides 160. If 5 divides $d$, then 25 divides 160, contradiction. So $d=1,2,4,8,16$ or 32.

If $d=1$, then $mn=250=2\cdot5^3$, but $m,n$ are coprime, so $a,b$ are $1,250$ or $2,125$. If $d=2$, then $mn=85$, so $a,b$ are $2,170$ or $10,34$.

If $d=4$, then 16 divides 360, contradiction. If $d=8$ then 64 divides 880, contradiction. If $d=16$, then 256 divides 1600, contradiction. If $d=32$, then $32^2$ divides 3040. Contradiction.

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  • $\begingroup$ can you explain it more clearly taking an example? $\endgroup$ – satyajeet jha Apr 26 '16 at 10:42
  • $\begingroup$ @satya Which part is unclear? $\endgroup$ – almagest Apr 26 '16 at 10:43
  • $\begingroup$ Hence d divides 160. If 5 divides d, then 25 divides 160, contradiction. So d=1,2,4,8,16 or 32. $\endgroup$ – satyajeet jha Apr 26 '16 at 10:46
  • $\begingroup$ We have $mnd^2-90d=160$, so $160=d(mnd-90)$. Hence $d$ and $mnd-90$ must be factors of 160. $\endgroup$ – almagest Apr 26 '16 at 10:47
  • $\begingroup$ the factor part was clear but i am not getting why you considered 5 divides 160 and then taking 25 doesn't ? $\endgroup$ – satyajeet jha Apr 26 '16 at 11:02

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