Mathematical proof?

Question

A) There do not exist three consecutive EVEN integers a, b, and c such that (a^2)+(b^2)=(c^2).
If true, prove it. Otherwise, give an counterexample.

B) There do not exist three consecutive ODD integers a, b, and c such that (a^2)+(b^2)=(c^2).
If true, prove it. Otherwise, give an counterexample.

Answer 1

As rainbowchaser92 mentioned, (6, 8, 10) is a counterexample to the first part.

For the second part, if a, b and c are any odd integers then a^2, b^2 and c^2 are odd. So a^2 + b^2 is even but c^2 is odd, so a^2 + b^2 ≠ c^2.

Answer 2

if you have 3 consecutive even integers, you may represent them as (2n, 2n+2, 2n+4). Then the above condition is
(2n)^2) + (2n+2)^2 = (2n+4)^2, or
4n^2 + 4n^2 + 8n + 4 = 4n^2 + 16n + 16, which can be simplified to
4n^2 - 8n - 12 = 0, or n^2 - 2n - 3 = = which has the solutions
n = (2 +/- sqrt(4 + 4*3) ) / (2) = 3 or -1
As a check, 6^2 + 8^2 = 10^2 is true and
(-2)^2 + 0^2 = 2^2 is also true.

So the assertion is false.

The assertion for odd integers gives:
(2n+1)^2 + (2n+3)^2 = (2n+5)^2, or
4n^2 + 4n + 1 + 4n^2 + 12n + 9 = 4n^2 + 20n + 25, or
4n^2 - 4n - 15 = 0 which has solutions
n = (4 +/- sqrt(16 - 4*4*(-15))) / (2*4)
= (4 +/- 16)/8 = -1.5 or 2.5

Since neither value for n are integers, the assertion is true.

Answer 3

A) I had a question like this in one of my math packets I had to do, the answer is yes.
In a 3, 4, 5 triangle you can double the sides to get a triangle with sides 6, 8, and 10
Now the triangle with sides 6, 8, 10 is obviously a right triangle. So 6^2+8^2=10^2
If we tried to find this alegebraicly, here is wha you get
x^2+(x+2)^2=(x+4)^2
x^2+x^2+4x+4=x^2+8x+16
x^2+4x+4=8x+16
x^2-4x-12=0
(x-6)(x+2)=0
x= 6 or -2
As you can see, you can't prove there are any odd consecutive integers like that. Possibly no.

Answer 4

A)

Pythagoras triplets are ideal to relate states of your question!

{(any odd number)^2} has 'an odd number digit' ending (1,5, and 9 alone)

A group of Pythagoras triplets do maintain a relation in between hypotenuse, x and y values as...

[{(any odd number)^2}/2]+/- 0.5 are two sides of a right angle triangle and said odd number is third side!

Example is (3^2/2)+/- 0.5 = 4.5 +/- 0.5 = 5 and 4

5^2 - 4^2 = 3^2, which is well known today!

(5^2 - 4^2 = 3^2) * 1
(5^2 - 4^2 = 3^2) * 2 ----->10^2 - 8^2= 6^2
(5^2 - 4^2 = 3^2) * 3 ------>15^2- 12^2=9^2 and so on!

It is obvious that triplet of all three sides as odd numbers will not exist as explained!


B)

There is another type of Pythagoras triplets that exists as groups ,which relates a circle

65^2= 63^2+16^2 or 60^2+25^2 or 52^2+39^2 or 56^2+33^2

Here {(even number ^2)/4} +/- 1 are Pythagoras triplets !

In this case 16^2= 256---> 256/4 = 64 & (65, 63, 16 is a pythagorus triplet)

65^2 - 63^2 =16^2 *1
65^2 - 63^2 =16^2 *2 ----> 130^2- 126^2 =32^2
65^2 - 63^2 =16^2 *3 ----->195^2- 189^2 =48^2

It is obvious that triplet of all three sides as even numbers exist as explained!

Edit :-

Last sentence was stated wrong as "It is obvious that triplet of all three sides as even numbers will not exist as explained!" which is now corrected by removing "will not" that was present in front of "exist"!


Regards!

Answer 5

A) Let the numbers be 2x, 2x+2, 2x+4.

To prove: (2x)^2 + (2x+2)^2 = (2x+4)^2
LHS
= 4x^2 + 4x^2 + 8x + 4
= 8x^2 + 8x + 4
RHS
= 4x^2 + 16x + 16

If LHS=RHS:
8x^2 + 8x + 4 = 4x^2 + 16x + 16
4x^2 - 8x - 12 = 0
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3 or -1
The possible numbers are 6,8,10 and -2,0,2.

B) Let the numbers be 2x+1, 2x+3, 2x+5.
To prove: (2x+1)^2 + (2x+3)^2 = (2x+5)^2
LHS
= 4x^2 + 4x + 1 + 4x^2 + 12x + 9
= 8x^2 + 16x + 10
RHS
= 4x^2 + 20x + 25

If LHS=RHS:
8x^2 + 16x + 10 = 4x^2 + 20x + 25
4x^2 - 4x - 15 = 0
(2x+3)(2x-5) = 0
x = -1.5, 2.5
The possible numbers are -2,0,2 and 6,8,10.
These are even numbers.
So there are NO odd integers that fulfil (a^2)+(b^2)=(c^2).

Answer 6

ummm A is false cause the pythagorean triple of 6,8,10

Answer 7

lol this is homework of course