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# If a, b, c are different positive integers, and a^2+b^2 = c^

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If a, b, c are different positive integers, and a^2+b^2 = c^  [#permalink]

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03 Aug 2013, 06:56
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If a, b, c are different positive integers, and a^2 + b^2 = c^2, then what is the value of (c - b)^2?

(1) a is a prime.

(2) b^2 is a multiple of 4.

Source: My own

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Re: If a, b, c are different positive integers, and a^2+b^2 = c^  [#permalink]

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Updated on: 03 Aug 2013, 08:21
3
mau5 wrote:
If a,b,c are different positive integers,and $$a^2+b^2 = c^2$$, then what is the value of $$(c-b)^2$$?

I.a is a prime.

II.$$b^2$$ is a multiple of 4.

$$a^2+b^2 = c^2$$
$$a^2= c^2-b^2 = (c-b)(c+b)$$

statement1:
$$a$$ is prime
therefore a = 1*a = (a cant have any other factor)
$$a^2= a*a = (c-b)(c+b)$$ ==>since a is prime $$c-b$$ should be $$1$$ and $$c+b$$ should be some perfect square.(as prime number factors are 1 and prime no.itself)
hence clearly $$c-b=1$$...hence sufficient

statement 2:
$$b^2$$ is a multiple of 4

let $$b=4 c=5 a=3$$....therefore $$c-b = 1$$
another case : let$$b=8 c=10 a=6$$...therefore $$c-b=2$$
hence insufficient

hence it is A
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Originally posted by blueseas on 03 Aug 2013, 08:15.
Last edited by blueseas on 03 Aug 2013, 08:21, edited 1 time in total.
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Re: If a, b, c are different positive integers, and a^2+b^2 = c^  [#permalink]

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03 Aug 2013, 08:16
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1
If a,b,c are different positive integers,and $$a^2+b^2 = c^2$$, then what is the value of $$(c-b)^2$$?
$$a^2=c^2-b^2=(c-b)(c+b)$$. Because $$c\neq{b},c-b\neq{0}$$ we get $$c+b=\frac{a^2}{c-b}$$.

I.a is a prime.
If a is prime, then $$a^2$$ has three factors: 1,a and $$a^2$$. Because c and b are integers, their sum must be an integer so $$c+b=integer=\frac{a^2}{c-b}$$. That is true if c-b is a factor of $$a^2$$:1, a or $$a^2$$. If $$c-b=a^2$$ then c+b=1 but this cannot be true as c and b are positive integers.
So we are left with $$c-b=a$$ or $$c-b=1$$. In the first case $$c-b=a$$, substituting in the above equation we find that $$(c-b)^2+b^2=c^2$$ or $$2b(b-c)=0$$, so b=0 (because $$c\neq{b}$$) but this contradicts the statement (b is a positive integer)
We are left with $$c-b=1$$, so $$(c-b)^2=1$$.
Sufficient

II.$$b^2$$ is a multiple of 4.
$$a^2+b^2 = c^2,3^2+4^2=5^2$$ and $$(c-b)^2=1$$
$$a^2+b^2 = c^2,(3*2)^2+(4*2)^2=(5*2)^2,=6^2+8^2=10^2$$ and $$(c-b)^2=4$$. Not sufficient.
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Re: If a, b, c are different positive integers, and a^2+b^2 = c^  [#permalink]

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25 Mar 2015, 12:10
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1
Hi All,

While the math involved in this question is certainly algebra-based, the patterns are rooted more in Geometry (specifically, the Pythagorean Theorem). You can use that knowledge to your advantage here.

We're told that A, B and C are DIFFERENT positive integers and that A^2 + B^2 = C^2. We're asked for the value of (C-B)^2.

Since the variables are INTEGERS, and if you recognize the Pythagorean Theorem, you might choose to jot down a few examples....

eg
3/4/5
5/12/13
And the multiples of those Triplets....

Fact 1: A is PRIME.

Right away, we have a couple of TESTs that we can run....

IF...
A = 3, B = 4, C = 5
The answer to the question is .... (C-B)^2 = (5-4)^2 = 1

IF...
A = 5, B = 12, C = 13
The answer to the question is .... (C-B)^2 = (13-12)^2 = 1

Since we only have two TESTs so far, I'd look to also TEST a rarer Pythagorean Triplet....

IF....
A = 7, B = 24, C = 25
The answer to the question is... (C-B)^2 = (25-24)^2 = 1

At this point, a pattern clearly exists. The answer to the question is ALWAYS 1.
Fact 1 is SUFFICIENT

Fact 2: B^2 is a multiple of 4

IF...
A = 3, B = 4, C = 5
The answer to the question is .... (C-B)^2 = (5-4)^2 = 1

IF...
A = 6, B = 8, C = 10
The answer to the question is .... (C-B)^2 = (10-8)^2 = 4
Fact 2 is INSUFFICIENT

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Re: If a, b, c are different positive integers, and a^2+b^2 = c^  [#permalink]

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09 Aug 2017, 18:49
a, b & c =+ve distinct integers
a²+b²=c²
or a²=c²-b²=(c-b)(c+b)
1) a is prime
plugin
a=3
a²=9
9=(c-b)(c+b)
now 9 is the product of (c-b)(c+b)
so it can be 1X9 or 9x1 or 3x3
c-b will be always less than c+b (sum of 2 positive distinct no will always be greater than their difference)
hence c-b can only be 1 and (c-b)²=(1)²-sufficient.
2) b² is multiple of 4
multiple values are possible
b²=4, 8, 16.....
b²=(c-a)(c+a)=1x4 or1x8 or 2x4 we can stop here-Not sufficient
Re: If a, b, c are different positive integers, and a^2+b^2 = c^   [#permalink] 09 Aug 2017, 18:49
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