2^(2x)-5^(2y)=39 I guess there is a reason for 2 showing up in powers so by diff of squares (2^x-5^y)(2^x+5^y)=1*39 or 3*13 because 2^x+5^y>0 always negative factors of 39 need not be considered. Also 2^x+5^y>2^x-5^y so we have only a few possible solutions 2^x-5^y=1 2^x+5^y=39 + => 2^x=20 no integer solutions Or 2^x-5^y=3 2^x+5^y=13 + => 2^x=8 x=3 and 2*5^y=10 y=1 So the only ineger solution (3,1)
Closed, analytical integer solutions are impossible for me. All I can do is guess to get the right answer, but I never know if it's unique.
What is the name of the program you use for writing?
I think he uses Notability. I think.
39 = 64 - 25 = 4^3 - 25^1 = 4^x - 25^y
x = 3 , y = 1
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2^(2x)-5^(2y)=39
I guess there is a reason for 2 showing up in powers so by diff of squares
(2^x-5^y)(2^x+5^y)=1*39 or 3*13 because 2^x+5^y>0 always negative factors of 39 need not be considered. Also 2^x+5^y>2^x-5^y so we have only a few possible solutions
2^x-5^y=1
2^x+5^y=39
+ => 2^x=20 no integer solutions
Or
2^x-5^y=3
2^x+5^y=13
+ => 2^x=8 x=3 and 2*5^y=10 y=1
So the only ineger solution (3,1)
2^×+5^y=13
&2^×-5^y=3
2(2^×)=16...=>×=3
&2(5^y)=10&y=1