This is a Pell equation with n=61 so it has solutions in integers. First we make the change of variables a=1+x and simplify, the equation becomes x(x+2)=61b2. Suppose 61 divides x so x=61ˉx which leads to 2+61ˉx=b2¯x
After substituting b=¯xm in (1) we find 2+61¯x=¯xm2 which implies that ¯x divides 2. Trying the possible values for ¯x leads to a contradiction.
After substituting ¯x=m2¯m and b=m¯m in (1) we find 2+61m2¯m=¯m which implies that ¯m divides 2. Trying the possible values for ¯m leads to a contradiction.
Now we can suppose 61 divides x+2. So x+2=61¯x. Substituting into the original equation gives 61¯x−2=b2¯x
I should say that this method only generates the trivial solution a=1 and b=0. But why doesn't it generate other solutions? What was the (wrong?) restrictive assumption that forced this method to generate only the trivial solution?
I also think this implies that the trivial solution is the only solution which shows that something must be wrong with this argument.
Answer
There is a problem at the point where I say that if $\bar{x} and ˉx divides b2 then ˉx divides b since ˉx may have a prime factor with power more than one whereas this prime number is present in b but raised to a smaller power while keeping the fact that ˉx is smaller than b.
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