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 $c^{2}-b{^2}$?

MathzSolver111 Nov 18, 2020

#1

#2**0 **

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

#3**0 **

b^{2} - a^{2} > 0 so b^{2} > a^{2} and so since a, b are positive, b > a

c^{2} - b^{2} > 0 so c^{2} > b^{2} 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:

b^{2} - a^{2} = 344

(a + 2)^{2} - a^{2} = 344

(a + 2)(a + 2) - a^{2} = 344

a^{2} + 4a + 4 - a^{2} = 344

4a + 4 = 344

4a = 340

a = 85

Since a = 85, b = 87 and c = 89

And so c^{2} - b^{2} = 89^{2} - 87^{2} = 352

hectictar Nov 18, 2020

#4**+1 **

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

#6**+1 **

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