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How many triples (a, b, c) of integers exists such that a^4 + b^3 = c^2

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How many triples (a, b, c) of integers exists such that a^4 + b^3 = c^2
posted Jun 17, 2019 by anonymous

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1 Answer

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a^4 + b^3 = c^2 --> c^2-a^4 = b^3 --> we can put it as (c-a^2)*(c+a^2)=b*b^2 -->
c-a^2=b--> a^2=c-b-----> (1)
c+a^2=b^2 -----> (2)
Considering (1) in (2) we have c+c-b=b^2 --> b^2+b-2c=0,
to solve it as quadratic equation, b=(-1+sqrt(1+8c))/2, it has infinite number of solutions concerning b&c
lets see what can a get, obviously it has much less solutions:
b=1--> c=1--> a=0
b=2--> c=3--> a=1
b=3--> c=6--> a^2=3, no integer solution
...
b=9--> c=45--> a=6
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b=50--> c=1275--> a=35
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b=289--> c=41905--> a=204
..
b=1682--> c=1415403--> a=1189
I got these numbers with the help of excel within first 3000 values of b.
So based on above there are infinite number of solutions

answer Jun 17, 2019 by Hanifa Mammadov



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