341(9,4) is the unique solution
Find all the pairs of positive numbers such that the last
digit of their sum is 3, their diference is a primer number and
their product is a perfect square.
(a previous olympiad question,interesting one)
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5 Answers
11If the numbers be a,b, with ab=c2 and a-b=p, then we have the following.....
a=\frac{(p+1)^2}{4}
b=\frac{(p-1)^2}{4}
c=\frac{p^2-1}{4}
Also 2p^2+2=10k+2 \Rightarrow p^2=5k.
From which we have p=5.
The unique triplet (a,b,c) follows.
341You know, u've got to give a lil bit of a roadmap as to why you conclude the relationship between a, b and p is as you have given.
11I thought that part was trivial.
Easy to see that (a,b)=1 or p....if (a,b)=p then we have k(k+1) as a perfect square, a contradiction.
Which means (a,b)=1, which also means a=m2 & b=n2.....
Now plugging in the fact that a-b=p, we have our desired relationship......
341well you could think this way: by giving a detailed solution, you could be helping someone who has never known this line of argument before.
