Question 1048316
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{{{1/4= 1/A + 1/B}}}

{{{AB = 4B + 4A}}}

{{{AB = 4(B+A)}}}

Since AB is a multiple of 4, there are two possibilities;

Case 1. One of them, A or B, is odd and the 
other a multiple of 4. 

Case 2. Both A and B are even

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Case 1: Say A is a multiple of 4 and B is odd

A = 4p,  B = 2q-1, for some positive integers p and q

{{{(4p)(2q-1)=4(2q-1+4p)}}}

{{{p(2q-1)=2q-1+4p}}}

{{{2pq-p=2q-1+4p}}}
{{{2pq-5p=2q-1}}}
{{{p(2q-5)=2q-1}}}
{{{p=(2q-1)/(2q-5)}}}
{{{p=(2q-5+5-1)/(2q-5)}}}
{{{p=(2q-5+4)/(2q-5)}}}
{{{p=(2q-5)/(2q+5)+4/(2q-5)}}}
{{{p=1+4/(2q-5)}}}

So 4 must be divisible by 2q-5

2q-5 = 1,2, or 4
  2q = 6,7, or 9

Since 2q is even, it can only be 6
  2q = 6
   q = 3
{{{p=1+4/(2q-5)}}}
{{{p=1+4/(2(3)-5)}}}
{{{p=1+4/(6-5)}}}
{{{p=1+4/1}}}
{{{p=1+4}}}
{{{p=5}}}

A = 4p,    B = 2q-1
A = 4(5)   B = 2(3)-1
A = 20     B = 6-1
A = 20     B = 5

Checking {{{1/A+1/B=1/20+1/5=1/20+4/20=5/20=1/4}}}

One solution is {A,B} = {20,5}

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Case 2:  Both A and B are even

A = 2p,  B = 2q, for some positive integers p and q

{{{AB = 4B + 4A}}}

{{{(2p)(2q)=4(2q)+4(2p)}}}

{{{4pq=8q+8p}}}

{{{pq=2q+2p}}}

{{{pq-2p=2q}}}
{{{p(q-2)=2q}}}
{{{p=2q/(q-2)}}}
{{{p=(2q-4+4)/(q-2)}}}
{{{p=(2q-4)/(q-2)+4/(q-2)}}}
{{{p=(2(q-2))/(q-2)+4/(q-2)}}}
{{{p=2+4/(q-2)}}}

So 4 must be divisible by q-2

q-2 =  1; 2; 4 
  q =  3; 4; 6 

Substitute in {{{p=2+4/(q-2)}}}

  p =  6; 4; 3
  A = 12; 8; 6 
  B =  6; 8;12

A and B are different, so that rules out the 8's,
so one is 6 and the other 12.

Checking: {{{1/6+1/12=2/12+1/12=3/12=1/4}}}

So there are two solutions:

{A,B} = {20,5}

and

{A,B} = {12,6}

Edwin</pre></b>