how-many-integer-a-b-z-a-5-b-5-10-b-1-2-9-

Question Number 174512 by ahdal last updated on 03/Aug/22

h o w m a n y i n t e g e r a, b \in z^{+}

a^{5} - b^{5} = 10 {(b + 1)}^{2} - 9

Answered by MJS_new last updated on 03/Aug/22

a^{5} = b^{5} + 10 b^{2} + 20 b + 1

b > 0 \Rightarrow a > b \Rightarrow a = b + n \land n \in N^{★}

{(b + n)}^{5} = b^{5} + 10 b^{2} + 20 b + 1

5 b^{4} n + 10 b^{3} n^{2} + 10 b^{2} n^{3} + 5 {b n}^{4} + n^{5} = 10 b^{2} + 20 b + 1

since b ⩾ 1 \land n ⩾ 1 {it}^{'} s obvious that the lhs is

greater than the rhs for higher values of

both b or n . the only solution is

b = 1 \land n = 1 \Rightarrow a = 2

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