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 Another case of divisibility (Posted on 2009-02-15)
Determine all pairs (a, b) of positive integers such that ab2 + b + 7 divides evenly a2b + a + b.

 See The Solution Submitted by pcbouhid No Rating

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 Analytic Solution | Comment 2 of 3 |

If a < b, then b > = a+1, giving:
a*(b^2) + b + 7 > a*(b^2) + b >= (a+1)(ab+1)
= (a^2)*b + a + ab > = (a^2)*b + a + b. Accordingly, no solution exists whenever a is less than b.

If possible, let us assume that a >= b

Let p = ((a^2)*b + a + b) / (a*(b^2) + b + 7)-------(#).

Accordingly, we obtain:

(a/b + 1/b)*(a*(b^2) + b+ 7)
= (a^2)*q + a + ab + 7*a/b + 7/b + 1
> (a^2)*b + a + b

So, p < a/b + 1/b.

Now, if b >=3, then (b - 7/b) > 0, and hence:
(a/q - 1/b)*(a*(b^2) + b + 7)
= (a^2)*b + a - a(b - 7/b) - 1 - 7/b
< (a^2)*b + a < (a^2)*b + a + b.

Hence, either b =1,2 ; or, p > a/b - 1/b
If, a/b - 1/b < p < a/b + 1/b; then, a-1 < bs < a+1.
Hence, a=b*s. Accordingly, from (#), we obtain:
(a,b)=(7*(p^2), 7p)

We now proceed to consider the b = 1,2

When, b = 1, then (a+8) divides (a^2)+ a + 1
Now, a(a+8) - ((a^2)+ a + 1) = 7*a -1,
also, 7(a+8) - (7*a - 1)= 57
The only factors of 57 greater than 8 are 19 and 57, so that :
a+ 8 = 19, 57 giving a = 11, 49; so that, (a,b) = (11,1), (49,1)

When, b = 2, then (4a+9) divides 2*(a^2)+ a + 2
Now, a(4*a + 9) - 2(2*(a^2)+ a + 2) = 7*a - 4,
also, 7(4*a+9) - 4(7*a - 4)= 79
The only factors of 79 greater than 9 is 79, so that:
4a+ 9 =79 giving a = 35/2, which is not an integer.
Hence, no positive integer solution is possible whenever q=2.

Summarizing, we have:
(a, b) = (11,1), (49,1), and (7*(p^2), 7p), whenever p is any given positive integer.

Edited on February 15, 2009, 3:49 pm
 Posted by K Sengupta on 2009-02-15 14:31:13

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