Find minimum of $a+b$ if $13|a+11b $ and $11|a+13b$ where $a,b>0$.
My attempt :
$13|a+11b \implies 13|a+24b$ . Similarly we get $11|a+24b$. Now $\gcd(11,13)=1$, so, $143|a+24b$.
Therefore $a+24b \geq 143$.
How to proceed after this?
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Here's a brute force approach: start with $a+11b=13k$ and $a+13b=11m$, with $k,m\in \mathbb N$. Solve it in $a,b$ and you get:
So $a$ and $b$ are integers iff $k$ and $m$ are both odd or both even; in this subspace for $(k,m)$, just minimize $78k-55m$.
It's 28. a+24b=n for all n has solutions since gcd(1,24)=1.Thus,a+24b=143z for all z has solutions in integers. For non- positive z obviously the equation cannot have solution in postive integers.Thus z is positive. For z=1,selecting the largest b such that 24b<143 we have b=5 and thus 23 + 24(5)= 143.For z>=5 we have 1+24(27)< 143(5)=715. Thus all equations where z>=5 have solutions such that a+b>28.