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Let's define five binomials as :

$P(a)=2a+1$

$Q(a)=3a+4$

$R(a)=4a+9$

$S(a)=5a+16$

$T(a)=6a+25$

How to prove that :

$\gcd(P(a),Q(a),R(a),S(a),T(a))=1$

for any particular value of $a$ , $(a\in \mathbb{Z^{+}})$ ?

I have checked statement by little Maple program for a many values of $a$:

enter image description here

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3  
We don't need all the conditions. If $p$ divides first two numbers, it divides $a+3$. If $p$ divides the next two numbers, it divides $a+5$. If it divides all three, it divides $a+5$ and $a+3$, so it divides $2$. But $p$ is not even by first condition. –  André Nicolas Nov 11 '11 at 8:13
    
Maple is a wonderful program, and Canadian too. But typing in the program seems like more work than the proof. –  André Nicolas Nov 11 '11 at 8:48
    
@Andre,right,but requires activation of less number of brain cells –  pedja Nov 11 '11 at 8:53
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2 Answers

up vote 2 down vote accepted

You only need the first three.

Any factor of each also divides a linear combination of them.

But $P(a)-a(P(a)+R(a)-2Q(a))= 2a+1-a(2a+1+4a+9-6a-8)=1$ so the highest common factor is 1.

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Actually, we only need the first two. :D –  user5137 Nov 11 '11 at 8:30
    
@Jack,that's not true..set $a=7$ –  pedja Nov 11 '11 at 8:32
    
@pedja - Oops! Right you are. The edit has been removed. –  user5137 Nov 11 '11 at 8:34
    
@Henry,result is $4a+1$ not $1$,you have a small typo,change sign in front of $a$ –  pedja Nov 11 '11 at 8:37
    
@pedja: thank you –  Henry Nov 11 '11 at 11:48
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We don't even need $T(a)$. Here's a proof that $gcd(P(a),Q(a),R(a),S(a))=1$.

Fix $a\in \mathbb{N}$ and suppose $d$ is a common divisor of each of $P(a)$, $Q(a)$, $R(a)$, and $S(a)$. Since $d\vert 2a+1$, we know that $d$ is odd.

Since $Q(a)=3a+4=(a+3) + (2a+1)$, it follows that $d \vert a+3$.

Since $R(a)=4a+9=(3a+4)+(a+5)$, it follows that $d \vert a+5$.

Further, $d$ divides $(a+3) + (a+5)=2a+8=(2a+1)+7$, hence $d \vert 7\,\Rightarrow\,d\in \{1,7\}$. Suppose $d=7$.

Then $7 \vert S(a)$, hence 7 divides $(4a+9)+(a+7)$. Thus, $7 \vert a+7$ and it follows that $7 \vert a$. However, then we would have $7=d \equiv 2a+1 \equiv 1 \mod(7)$, a contradiction.

Therefore the only common divisor of $P(a)$, $Q(a)$, $R(a)$, and $S(a)$ is 1 and hence $gcd(P(a),Q(a),R(a),S(a))=1$.

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1  
....or you can get a shorter version of this via Andre's comment above. Well done! –  user5137 Nov 11 '11 at 8:18
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