Math Challenge

Mar 2018
18
0
California
Hey y’all,
See if you can solve this problem:
Find all $(a,b)$ of positive integers such that $a^3 - 6b^2 =2018$ and $b^3 - 6a^2 = 155$ hold simultaneously.

Also don’t just post an answer, justify them with solutions.
 

greg1313

Forum Staff
Oct 2008
8,008
1,174
London, Ontario, Canada - The Forest City
The difference between the two equations is

$$(a-b)(a^2+ab+b^2+6a+6b)=1863$$

$1863$ factors as $3^4\cdot23$ and its divisors are $1,3,9,23,27,69,81,207,621,1863$ so $a-b$ is one of $1,3,9,23,27$. Using this we can solve for $a$ in terms of $b$ and write the quadratic in our original factorization purely in terms of $b$. If it has a positive integer root for $b$ then that root is a solution. Applying this we find the quadratic has a positive integer root for $a=b+3$, specifically $b=11$.

All $(a,b)$ of positive integers such that $a^3−6b^2=2018$ and $b^3−6a^2=155$ hold simultaneously are $(14,11)$.
 
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Denis

Math Team
Oct 2011
14,592
1,026
Ottawa Ontario, Canada
Evident that a is even and b is odd.

a^3 - 6b^2 = 2018 [1]
-6a^2 + b^3 = 155 [2]

[1] * b + [2] * 6 leads to b = (36a^2 + 930) / (a^3 - 2018 )

a^3 - 2018: a > 12 (so 14, 16 ....)

36a^2 + 930: multiple of 14, 16 ....

Only (a,b) = (14,11)

Above is "screwing around" !!
 
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Country Boy

Math Team
Jan 2015
3,261
899
Alabama
Evident that a is even and b is odd.

a^3 - 6b^2 = 2018 [1]
-6a^2 + b^3 = 155 [2]

[1] * b + [2] * 6 leads to b = (36a^2 + 930) / (a^3 - 2018 )

a^3 - 2018: a > 12 (so 14, 16 ....)

36a^2 + 930: multiple of 14, 16 ....

Only (a,b) = (14,11)

Above is "screwing around" !!
An old and well respected method of solving math problems!
 
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