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Solve the Diophantine Equation 36^a - 4^b = 272
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[RekhaSharma-eose]

I had solved almost your questions and I really likes your questions

[VictorLonmo]

Another approach is to recognize that 36^a can be written as (4^a)(9^a) so the right hand side becomes (4^a)(9^a) -(4^b). Define x = b-a. Rewrite the right hand side as (4^a)[(9^a) -(4^x)]. Then factor the left hand side so 272 becomes (16)(17).
At that point it becomes clear that a must equal 2.
Divide both sides by (4^2) aka 16
The equation becomes: (9^2) -(4^x) = 17
or 81 -(4^x) = 17
rearrange this equation for 4^x and we get 4^x = 81-17
or 4^x = 64
therefore x = 3
recall x = b-a
this becomes (3) = b - (2)
or b = 5
This solves the question without the need for trying different answers to get to the solution.

Thanks to PreMath for the videos!

[roger]

I solved this problem in my head knowing that a and b would be integers. 36^2 = 1296, 1296 - 272 = 1024, 4^5 = 1024. Thus a = 2 and b = 5.

[ralkadde]

There is no need to check all factors in detail. Looking at (6^a-2^b) * (6^a+2^b), the right factor is 2^(b+1) greater than the left one. So we restrict our searching to those factors that have a power of 2 as difference. 136-2=134 failed, 68-4=64 okay, 34-8=26 failed. 4*68 is the only factor that works. That leads to 6^a-2^b = 4 and 6^a+2^b = 68.

[kennethstevenson]

Since 36 raised to a power greater than 2 would be difficult to be decreased by a power of four to produce 272. So I squared 36 and got 1296 subtracted 272 and got 1024 which I recognized as a power of 4 (4^5) Therefore a=2 and b=5.

[heenaandrobin]

I done it with another way and it is also correct, , , I did this question in only 5 to 6 taking 4^a common and then equating by doing factor os 272 as 2^4×17

[davidfromstow]

Incredibly interesting and informative - really enjoyed it.

[yalchingedikgedik]

Thats very nice and useful.
We are learning from you .
Thanks Sir .

[dgpztur]

🌿🌺🌿Καλό μεσημέρι...🌿🌺🌿 Απ' ΚΡΉΤΗ 🌿 🌺🌿

[davidbrisbane]

First we note that neither a nor b can be negative, as otherwise the LHS of the equation is a fraction and the RHS of the equation is a whole number.

So, moving forwards, let's consider the case where a and b can be zero, or positive integers.

Clearly, both a and b can't be zero, as then the LHS of the equation would be zero and not equal to 272.

One of a or b can't be zero either, as then the LHS of the equation would be odd and the RHS of the equation would be even.

So, we are forced to conclude, that if the equation has solutions, then a and b must be positive integers.

Case (a = b)
36ᵃ - 4ᵇ = 36ᵇ - 4ᵇ = 4ᵇ (9ᵇ - 1) = 272 = 16 × 17, where 17 is the only odd factor of 272. This being the case, there is only a solution if 9ᵇ - 1 = 17, and this has no solution in positive integers for b. So, a can't be equal to a.

Case(a > b)
Let a = b + h, where h is a positive integer.
So, 36ᵃ - 4ᵇ = 9ᵇ*4ᵇ*9ʰ*4ʰ - 4ᵇ =
4ᵇ(9ᵇ*4ᵇ*9ʰ - 1) =
272 = 16 x 17.

Now (9ᵇ*4ᵇ*9ʰ - 1) is odd, so it must equal the only odd factor in 272, ie. 17.

So, 9ᵇ*4ᵇ*9ʰ - 1 = 17

Clearly, there is no positive b and h integers that satisfy this condition, so a can't be larger than b.

Case(a < b)
Let b = a + h, where h is a positive integer.
So, 36ᵃ - 4ᵇ = 9ᵃ*4ᵃ - 4ᵃ*4ʰ =
4ᵃ(9ᵃ - 4ʰ) = 272.

Now 9ᵃ - 4ʰ = 17, as 17 is the only odd factor in 272 and 4ᵃ must divide the other factor of 272, namely 16. So, a = 1, or a = 2.

But a = 1 doesn't solve 9ᵃ - 4ʰ = 17 for any possible integer h.

Now when a = 2, we find 9ᵃ - 4ʰ = 9² - 4ʰ = 17.
So, 4ʰ = 81 - 17 = 64.
So, h = 3 is a solution, which means that b = a + h = 2 + 3 = 5.

So, (a, b) = (2, 5) is the only solution to 36ᵃ - 4ᵇ = 272 in integers a and b.

[vaibhavgupta]

Did you learned how to solve these questions in your school 🎒

[aimanusmani]

Which software you are using I have to prepare my assignment

[mikejames]

simpler solution (4x9)^a -4^b = 4x68 4^a x9^a -4^b=4x68; 4^a-2 x9^a -4^b-2 =17; 4^a-2 x9^a = 4^b-2 +17; If a not=2 LHS is even and a multiple of 4; if b not=2 RHS is odd (rejected), If b=2 RHS =18, LHS is not multiple of 4 (rejected). If a =2 LHS = 4^0 x 9^2 =81; 81--17 = 64= 4^3= 4^b-2 Therefore b-2 =3 and b=5 Solution

[mariusk]

I do not understand why the first (1 x 272) and the last (16 x 17) were rejected

[notunkori]

If x^3-1/x^3=22sqr2 then what will be the value of x^2+1/x^2=?
Sir I badly need this solution. So publish very soon

[giuseppemalaguti]

modo per essere l'argomento del log4 pari(EVEN) é a=2 che dà b=5

[math-physics]

Have you got the question as you told me? By the way today's question is something