Counting integer solutions in a nested radical equation

Expert intro: Nested radical equations like $\sqrt{x-\sqrt{x+23}}=2\sqrt{2}-y$ require matching both rational and irrational parts carefully.

Detailed walkthrough

Step 1: Square the structure, not just the symbols

The equation is $\sqrt{x-\sqrt{x+23}}=2\sqrt2-y,$ with $x,y\in\mathbb Z$. Because the left-hand side is a square root, it is always nonnegative, so the right-hand side must also be nonnegative. More importantly, the left side has the form “integer minus a square root of an integer,” while the right side is “integer minus $2\sqrt2$.” That tells us the irrational part must match exactly.

Square both sides: $x-\sqrt{x+23}=(2\sqrt2-y)^2=y^2+8-4y\sqrt2.$ Now compare rational and irrational parts. The left side has rational part $x$ and irrational part $-\sqrt{x+23}$. The right side has rational part $y^2+8$ and irrational part $-4y\sqrt2$.

Step 2: Match the rational and irrational parts

For these two expressions to be equal, we need $x=y^2+8$ and $\sqrt{x+23}=4y\sqrt2.$ The second equation implies $x+23=32y^2.$ Substitute $x=y^2+8$ into this: $y^2+8+23=32y^2,$ so $31=31y^2,$ which gives $y^2=1.$ Thus $y=\pm1$.

Now check which sign works in the original equation. If $y=-1$, then the right side is $2\sqrt2+1$, whose square has a positive irrational part. But the left side becomes $x-\sqrt{x+23}=9-4\sqrt2$ when $x=9$, which is too small to equal $(2\sqrt2+1)^2$. So $y=-1$ does not work.

If $y=1$, then $x=9$, and the equation becomes $\sqrt{9-\sqrt{32}}=2\sqrt2-1.$ Since $\sqrt{32}=4\sqrt2$, the left side is $\sqrt{9-4\sqrt2},$ and indeed $(2\sqrt2-1)^2=8+1-4\sqrt2=9-4\sqrt2.$ So $x=9$, $y=1$ is a valid solution.

Step 3: Count the pairs

There is exactly one integer pair $(x,y)$ that works.

The correct choice is B. 1.

Common pitfall

A frequent mistake is to square immediately and then treat $\sqrt{x+23}$ as if it could be any real number independent of $x$. It cannot: because $x$ is an integer, the radical part has to fit an integer-based structure. Another common error is forgetting that the right side must be nonnegative before squaring. If $2\sqrt2-y<0$, then no solution is possible because a square root cannot equal a negative number.

The cleanest path is to square once, match the rational and irrational parts, and then verify the surviving candidate directly in the original equation.

💡 Pitfall guide

The difficult step is not the algebra of squaring; it is recognizing that $x-\sqrt{x+23}$ and $2\sqrt2-y$ live in different number types unless their irrational parts line up perfectly. If you jump straight to expanding both sides without comparing the rational and irrational components, you may keep impossible candidates. Another trap is accepting both $y=1$ and $y=-1$ after solving $y^2=1$. The original equation is asymmetric because the right side is $2\sqrt2-y$, so the sign matters. You must check any candidate back in the unsquared equation, not just in the squared one.

🔄 Real-world variant

If the equation were changed to $\sqrt{x-\sqrt{x+23}}=3\sqrt2-y$, the same method would still apply, but the matching equations would change. After squaring, the right side becomes $y^2+18-6y\sqrt2$, so you would compare it with $x-\sqrt{x+23}$. That leads to $x=y^2+18$ and $\sqrt{x+23}=6y\sqrt2$, hence $x+23=72y^2$. Substituting gives $y^2+41=72y^2$, so $71=71y^2$ and again $y=\pm1$. Verifying in the original equation would then determine which sign works. This variant shows how the coefficient of $\sqrt2$ changes the entire solution count even though the method stays the same.

🔍 Related terms

nested radical equation, rational and irrational parts, integer solution count