Math Olympiad | How to Find Integer Solutions to the Exponential Equation?

This math tutorial video shows you how to find integer solutions to an exponential exponential equation: 3^(2x)-2^(2y)=77. Tricks such as exponent rules, difference of squares formula, and prime factorization of integers are used. The techniques used here are useful for solving other types of equations such as Diophantine equations.

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[DrLiangMath]

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[에스피-ht]

Solution by insight
81-4=77
3^2x=81, x=2
2^2y=4, y=1

[deomanisharmaramsurrun]

Thanks so much, Sir.
Excellent and detailed explanation.
It was a joy to listen.

[owlsmath]

Nice problem! Enjoyed the explanation

[thomashammond]

Man's really mastered the whole title and thumbnail thing

[peterhofer]

I was wondering what happens if you allow real numbers for x, y.

On can use again the idea

a × b = 77
a = 3^x + 2^x
b = 3^x - 2^x
With the restriction a > b

y = log( (a^2 - 77) / 2a ) / log 2
x = log( (77/a + 2^y) ) / log 3

You get a function of all valid pairs (a, b) element (R × R). So you can choose e.g. some a.

With a=50, you get
b = 77/50 = 1.54
y = 4.5987224
x = 2.957559

And (3^x + 2^y)(3^x - 2^y) = 49.9999 × 1.53999 ca= 77

It's nice to see that this works out.

[mohamedabdelkaderahmed]

@
Math Tutoring by Dr. Liang
Hello Dr.Liang, , , I've solved it in a different way & I need your comment

Let 3^x=a & 2^y=b
a^2-b^2=77
a^2=77+b^2

As we have only one equation of two variables, so I will assume the solution as follows:-

a^2 must be greater than 77 so if we try a=9 so a^2=81
so b^2=81-77=4, b=2
Now, 9=3^x, so x=2, 2^y=2 , so y=1