A²,b²,c² are consecutive perfect squares.How many natura numbers are lying between a² and c², if a>0

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A²,b²,c² are consecutive perfect squares.How many natura numbers are lying between a² and c², if a>0

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Nho 4 years 2021-07-26T08:16:41+00:00 1 Answers 7 views 0

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  1. minhkhang
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    2021-07-26T08:17:47+00:00
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    Answer:

    The quantity of natural numbers between a^{2} and c^{2} is 2\cdot (a + b) + 1.

    Step-by-step explanation:

    If a^{2}, b^{2} and c^{2} are consecutive perfect squares, then both a, b and c are natural numbers and we have the following quantities of natural numbers:

    Between b^{2} and c^{2}:

    c^{2} = (b+1)^{2}

    c^{2} = b^{2}+2\cdot b + 1

    c^{2}-b^{2} = 2\cdot b + 1

    And the quantity of natural numbers between b^{2} and c^{2} is:

    c^{2}-b^{2}-1 = 2\cdot b

    Between a^{2} and b^{2}:

    b^{2} = (a + 1)^{2}

    b^{2} = a^{2} +2\cdot a + 1

    b^{2}-a^{2} = 2\cdot a + 1

    And the quantity of natural numbers between a^{2} and b^{2} is:

    b^{2}-a^{2}-1 = 2\cdot a

    And the quantity of natural numbers between a^{2} and c^{2} is:

    Diff = 2\cdot a + 2\cdot b + 1

    Please observe that the component +1 represents the natural number b^{2}

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