[sp]anemone said:Find all the natural numbers $a$ and $b$ such that $(a^2+1)(b^2+1)+45=2(2a+1)(3b+1)$.
Opalg said:[sp]
Rewriting the equation: $$a^2b^2 + a^2 + b^2 + 46 = 12ab + 4a + 6b + 2,$$ $$a^2b^2 -12ab + a^2 - 4a + b^2 - 6b = -44,$$ $$(ab-6)^2 + (a-2)^2 + (b-3)^2 = -44 + 36 + 4 + 9 = 5.$$ The only way to express $5$ as the sum of three squares of integers is $4+1+0$. Therefore one of the three squares in the sum $(ab-6)^2 + (a-2)^2 + (b-3)^2$ must be $0$. Taking the three cases in turn, we must have either $ab=6$ or $a=2$ or $b=3$.
If $a=2$ the equation can be satisfied by taking $b=2$ or $b=4$. The other two cases $ab=6$ and $b=3$ do not lead to integer solutions. Therefore the only solutions for $(a,b)$ are $(2,2)$ and $(2,4).$
[/sp]
Natural numbers are the set of positive integers (1, 2, 3, 4, ...), starting from 1 and increasing by 1 indefinitely.
To solve an equation in the set of natural numbers, you must find a value for the variable that satisfies the equation and is also a natural number. This can be done by using algebraic operations to isolate the variable on one side of the equation.
No, not all equations can be solved in the set of natural numbers. Some equations may have solutions that are not natural numbers, such as fractions or negative numbers.
Yes, there are some strategies that can be helpful when solving equations in the set of natural numbers. These include using substitution, trial and error, or setting up a system of equations to solve for multiple variables.
Yes, equations in the set of natural numbers can have multiple solutions. This is because there may be more than one natural number that satisfies the equation. It is important to check all possible solutions when solving equations in the set of natural numbers.