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balandron [24]
3 years ago
11

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

Mathematics
1 answer:
pogonyaev3 years ago
4 0

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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