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right that is what I have so far

ok lets subtract the two equations

(b+3) - b = mb - (a (k+1))

so subtract b+3-b?

yes

yes

system of equations

i get 3 = mb - ak - a

that is what I got too

is there more info
are a and b must be integers?
well yes if we are discussing "divides"

ok now substitute back

suppose a and b are positive integers

nevermind, thats circular to substitute back

wait I have a question

isn't b= (a+1)k? and b+3= bj?

i have a different idea

yes , thats correct

add 3 to the first equation , so b+3 = 3 +(a+1)k

so bj = 3 + (a+1)k

so bj - 3 = (a+1)k

why do you have (b+3)-b= mb-(a(k+1))?

wait you did that backwards

isn't it suppose to be mb-(a+1)k?

b= (a+1)k? and b+3= bj? this is false

if a+1 divides b then b=(a+1)k?

isnt the definition of divisiblity if a divides b then b= ak?

ok so we have bj-3 = (a +1) k ,

yes i misread

how did you get bj-3= (a+1)k?

a few steps

ok first step, b = (a+1)k , b+3 = bj

did you do that by substituting?
i see what you did

3 = bj - (a+1)k

rearranging we get

(a+1)k = bj - 3

but we know that b = (a+1) k

so b = bj - 3

follow so far, i might have gone too fast

b = (a+1)k , b+3 = bj , subtracting the former from the latter we get
3 = bj - (a+1)k

add (a+1)k to both sides, subtract 3 from both sides
you should get bj = (a+1)k ,

I got it so far

woops

bj - 3 = (a+1)k

ok we know that (a+1)k = b, so by transitive rule we have
b = bj - 3

so b+3 = bj , so b divides b+3, which we already know, shoot

ok lets do some substitution

b = (a+1)k , b+3 = bj ,
thats the given.
substitute (a+1)k for b in the second equation

so we get (a+1)k + 3 = (a+1)k*j

rewriting it as b= bj-3 does that mean that we proved b=3 since that can be rewritten as b-3=0?

how did you get that? i got b + 3 = bj

ok i have the answer

bj-3= (a+1) k and we know that b=(a+1)k

so b = bj - 3

which is b + 3 = bj , and we already know this

ok then, ready for the solution

yeah but I am not sure if that proves that b=3?

it doesnt add any new information
its not helpful

so we abandon that

different approach

right

good
now divide both sides by a+1

just to make sure we are starting all over?

yes

did you get a+1)k + 3 = (a+1)k*j

i got k+ 3/(a+1)= kj

so 3/(a+1) = kj - k

so 3/(a+1) is an integer

because kj and k are integers, and the difference of integers is always an integer

or 3/(a+1) = k ( j-1)

there is only one positive value that will make 3/(a+1) an integer

is that wrong? I did it twice and that is what i got

a=2 will make 3/(a+1) = 3/3 = 1

or backtrack

we assumed, k, j, and a are positive initially

we can prove this actually

, k, j, a, b are all positive

ok so far?

we'll go back to show that k and j are positive, but lets just assume it is at the moment

so we have k + 3/ (a+1) = kj, agreed?

right but what does the k(j-1) mean in this context?

well we dont need that. that just shows the left left is an integer

but i want to make this airtight

to avoid the negative cases

k + 3/(a+1) = kj , agreed?

dont subtract k from both sides

right hand side is positive, and left hand side is positive

Yes. so we would have 2 cases where k and j are even and then another where k and j are odd

dont need that

for this to be an equality, then we need kj to be a positive integer

why?

no, it isnt always positive

if j = 1, 1 -1 = 0

we dont have to subtract k from both sides

and since k>0 , j>0 , we have k>0 , j>1

ok we can do that

we know that k(j-1) is an integer, correct? since integers are closed under multiplication

ok so far?

you still there?

yes

yes im here

Ok I get this a little bit more but then how would we do this for b=3 ?

then you just plug it back in

then we can use our other formula

you mean it won't work when a=1 or a=3 because we get fractions, which aren't integers?

right

which other formula?

3 = bj - 2k

after substituting a = 1

wait shouldn't it be a=2 not 1?

oh

ok

so substituting we get

b = 3k for the first equation

since b = (a+1)k

agreed?

you there?

gotcha

yes

i have a much shorter proof

but going with this

substituting 3k for b into second equation we get

3k + 3 = 3kj

right

so 3 = 3kj - 3k

3 = 3k ( j - 1)

yes

so 1 = k ( j-1)

yup

1/ j-1 = k, and we know that k > 0 and positive integer

so j = 2, k = 1

so now we had before, b + 3 = bj,
so b + 3 = b*2
so b = 3

right

here is a shorter proof

would we put this as 2 two cases or just a long direct proof?

that was a direct proof

here is a shorter direct proof

that was excruciating long :)

b+3 = bj, so 3 = bj - b
or 3 = b (j-1), so 3 is divisible by b

so since b > 0 , there is only 2 options for b , b = 1 or b = 3

ok so far?

yes but b would be 0 if its 1

only 1 goes into 3 evenly and 3 goes into 3 evenly

what do you mean?

oh nevermind

you mean a = 0 if b = 1
, right

i was thinking about j

we have b = (a+1) k from second equation

ok

wait thats false what you said

if b = 1 , we have b+3 = bj , so 1 + 3 = 1(4) or j = 4

and if b = 2 , we have b + 3 = b j , so 3 + 3 = 3(2)

err
, if b = 3 i mean

the second equation clinches the deal , b = (a+1)k , k = b / (a+1)

k = 1/ (a+1) has only a = 0 solution if k is an integer

but we assumed that a > 0 , so contradiction. therefore b = 3

your welcome
i feel stupid now that i did such a long proof

always look for a trick, to avoid time loss