Solve in the natural numbers the equation
9x-3x=y4+2y3+y2+2y
Factor both sides
3^x (3^x -1) = y(y+2)(y^2+1)
Observe 3^x has factors of 3 so either
Case 1 y is divisible by 3: y=0mod3
Case 2 y+2 is divisible by 3: y=1mod3
(y^2+1) is never divisible by 3
(3^x-1) is never divisible by 3.
Case 1:
y=3^x
and
3^x-1 = (y+2)(y^2+1)
A quick graph shows this system has no solutions
Case 2
y+2=3^x
and
3^x-1=y(y^2+1)
A quick graph shows these two curves cross only at (1,1)
https://www.desmos.com/calculator/orsd08h2ph
You could prove the above cases with higher math than I feel like doing. The graphs show pretty clearly.
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Posted by Jer
on 2021-02-12 11:10:29 |