Product of a 97th Power and a 997th Power

Number Theory Level 4

What is the minimum positive integer value of $N$ satisfying the condition below?

For every ordered pair of positive integers $(m,n)$ (with $n\geq N$ ), there are positive integers $a$ and $b$ with $\large m^n=a^{97}b^{997}.$

The answer is 95616.

1 solution

Kushal Bose
Jan 2, 2017

Let $n=97x+997y$ for $x,y$ are non-negative integers then $m^n=m^{97x+997y}=(m^x)^{97} (m^y)^{997}=a^{97} b^{997}$ where $m^x=a$ and $m^y=b$

So, we have to find out largest number $n=97x+997y$ where there will be no solution

This is a problem of Frobenius Number .Here $gcd(97,997)=1$ So ,largest Frobenius number is $97 \times 997 -97 -997=95615$

So, if $n=95615+1=95616$ then there will be a solution in $(a,b)$

This is the right solution. But, I think, a bit incomplete. Just add the condition that both $x$ and $y$ are non-negative integers.

And what do you think about this matter: "we also have to show that--- if $n= 97x+997y$ has no solution in non-negative integers $x$ and $y$ , then NOT for every $m$ , there exist positive integers $a$ and $b$ such that $m^n=a^{97}b^{997}$ "?

Muhammad Rasel Parvej - 4 years, 5 months ago

Product of a 97th Power and a 997th Power

The answer is 95616.

1 solution

0 pending reports