The equation \((a^2+1)(b^2+1)+45=2(2a+1)(3b+1)\) has been solved for natural numbers \(a\) and \(b\). The rewritten form leads to the equation \((ab-6)^2 + (a-2)^2 + (b-3)^2 = 5\). The only valid solutions occur when \(a=2\), yielding \(b=2\) or \(b=4\). Thus, the complete set of solutions is \((2,2)\) and \((2,4)\).
PREREQUISITESMathematicians, educators, and students interested in number theory, particularly those focusing on solving equations in the set of natural numbers.
[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 $(2,4).$
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