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# I would kindly request some help.

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Three consecutive positive odd integers \$a,b\$ and \$c\$ satisfy \$b^{2}-a^{2}=344\$ and \$c^{2}-b^{2}>0.\$ What is the value of \$c^{2}-b{^2}\$?

Nov 18, 2020

#1
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Using algebra, c^2 - b^2 = 348.

Nov 18, 2020
#2
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No explanation? Just "use algebra"? Thanks for trying.. but I need a good explanation I'm not here just for a quick answer.

MathzSolver111  Nov 18, 2020
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b2 - a2 > 0   so  b2 > a2  and so since a, b are positive,  b > a

c2 - b2 > 0   so  c2 > b2  and so since c, b are positive,   c > b

So   a < b < c

And since a, b, and c are consecutive positive odd integers:

b  =  a + 2

c  =  a + 4

Now we can find  a  using this equation:

b2 - a2  =  344

(a + 2)2  -  a2  =  344

(a + 2)(a + 2) - a2  =  344

a2 + 4a + 4 - a2  =  344

4a + 4  =  344

4a  =  340

a  =  85

Since  a = 85,  b = 87  and  c = 89

And so   c2 - b2  =  892 - 872  =  352

Nov 18, 2020
#4
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Thanks hectictar! Also, happy \$21\$st and \$22\$nd birthday (sorry I couldn't say this at your real birthday I joined two weeks ago).

MathzSolver111  Nov 18, 2020
edited by MathzSolver111  Nov 18, 2020
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Thanks! xD

hectictar  Nov 18, 2020
#6
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The answer was correct and the exact explanation for how to solve after I put in \$352\$ was almost exactly what you said! Thank you a million!

MathzSolver111  Nov 18, 2020