Kvant Math Problem 1522

Let $x=\sqrt{m}$ and $y=\sqrt{m+d}$.

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Verified: yes
Verdicts: PASS + PASS
Solve time: 32m23s
Source on kvant.digital

Problem

Prove that for any natural numbers $m$, $d$, $k$ there exists a natural number $n$ such that $$\left(\sqrt{m}+\sqrt{m+d}\right)^k=\sqrt{n}+\sqrt{n+d^k}.$$

P. Filevich

Exploration

Let $x=\sqrt{m}$ and $y=\sqrt{m+d}$. Then $x^2=m$, $y^2=m+d$, and

$y^2-x^2=d.$

Hence

$(x+y)(y-x)=d.$

Set

$S=(x+y)^k,\qquad T=(x-y)^k.$

Then

$ST=(x+y)^k(x-y)^k=(x^2-y^2)^k=(-d)^k.$

The desired representation suggests introducing

$A=\frac{S+T}{2},\qquad B=\frac{S-T}{2}.$

If $A^2$ and $B^2$ are natural numbers and if $B^2=A^2+d^k$, then

$S=A+B=\sqrt{A^2}+\sqrt{A^2+d^k},$

which gives the required form with $n=A^2$.

The key point is to prove that $A$ and $B$ are themselves square roots of natural numbers, namely that $A^2$ and $B^2$ are integers.

Problem Understanding

This is a Type D existence problem. One must construct a natural number $n$ such that

$(\sqrt m+\sqrt{m+d})^k=\sqrt n+\sqrt{n+d^k}.$

The natural construction is

$n=\left(\frac{S+T}{2}\right)^2,$

where

$S=(x+y)^k,\qquad T=(x-y)^k.$

The remaining work is to prove that this $n$ is a natural number and that

$\left(\frac{S-T}{2}\right)^2=n+d^k.$

Proof Architecture

Introduce $x=\sqrt m$, $y=\sqrt{m+d}$ and define

$S=(x+y)^k,\qquad T=(x-y)^k.$

The first step is to show that $A=(S+T)/2$ and $B=(S-T)/2$ have squares that are integers. This comes from the parity pattern in the binomial expansions.

The second step is to compute

$B^2-A^2=-ST.$

Using

$ST=(-d)^k$

and the fact that the sign of $T$ depends on the parity of $k$, one obtains

$B^2=A^2+d^k$

or

$A^2=B^2+d^k.$

Then

$S=A+B=\sqrt{A^2}+\sqrt{A^2+d^k},$

or the analogous identity with $A$ and $B$ interchanged.

Solution

Let

$x=\sqrt m,\qquad y=\sqrt{m+d},$

and define

$S=(x+y)^k,\qquad T=(x-y)^k.$

Expanding by the binomial theorem,

T=\sum_{j=0}^{k}\binom{k}{j}x^{k-j}(-y)^j.$$Adding,$$S+T=\sum_{j=0}^{k}\binom{k}{j}x^{k-j}y^j(1+(-1)^j).$$Only even indices $j=2r$ remain, so$$S+T= 2\sum_{r=0}^{\lfloor k/2\rfloor} \binom{k}{2r}x^{k-2r}y^{2r}.$$If $k$ is even, then $k-2r$ is even for every $r$. Writing$$x^{k-2r}=(x^2)^{(k-2r)/2}=m^{(k-2r)/2},$$and$$y^{2r}=(y^2)^r=(m+d)^r,$$each summand is an integer. Hence$$A=\frac{S+T}{2}\in\mathbb Z.$$If $k$ is odd, then $k-2r$ is odd. Writing$$x^{k-2r} =x(x^2)^{(k-2r-1)/2} =x,m^{(k-2r-1)/2},$$gives$$A= x\sum_{r=0}^{(k-1)/2} \binom{k}{2r} m^{(k-2r-1)/2} (m+d)^r.$$Thus$$A=xM$$for some integer $M$, and therefore$$A^2=x^2M^2=mM^2\in\mathbb Z.$$In both parity cases, $A^2$ is a natural number. Performing the same calculation for $S-T$,$$S-T = \sum_{j=0}^{k} \binom{k}{j} x^{k-j}y^j(1-(-1)^j),$$so only odd indices $j=2r+1$ survive:$$S-T= 2\sum_{r=0}^{\lfloor (k-1)/2\rfloor} \binom{k}{2r+1} x^{k-2r-1}y^{2r+1}.$$If $k$ is even, then $k-2r-1$ is odd. Writing$$x^{k-2r-1} = x(x^2)^{(k-2r-2)/2} = x,m^{(k-2r-2)/2},$$and$$y^{2r+1} = y(y^2)^r = y(m+d)^r,$$gives$$B = xy \sum_r \binom{k}{2r+1} m^{(k-2r-2)/2} (m+d)^r.$$Hence$$B=xyN$$for some integer $N$, and$$B^2=x^2y^2N^2=m(m+d)N^2\in\mathbb Z.$$If $k$ is odd, then $k-2r-1$ is even. Writing$$x^{k-2r-1} = (x^2)^{(k-2r-1)/2} = m^{(k-2r-1)/2},$$and$$y^{2r+1} = y(y^2)^r = y(m+d)^r,$$each term of $B$ contains a factor $y$. Hence$$B = y\sum_r \binom{k}{2r+1} m^{(k-2r-1)/2} (m+d)^r.$$Therefore$$B=yN$$for some integer $N$. For example, when $k=1$ this gives $B=y$, which matches the formula above. Consequently,$$B^2 = y^2N^2 = (m+d)N^2 \in\mathbb Z.$$Hence $B^2$ is a natural number in all cases. Define$$n=A^2=\left(\frac{S+T}{2}\right)^2.$$Now compute$$ST=(x+y)^k(x-y)^k=(x^2-y^2)^k=(-d)^k.$$Also,$$B^2-A^2 = \left(\frac{S-T}{2}\right)^2 - \left(\frac{S+T}{2}\right)^2.$$Expanding,$$B^2-A^2 = \frac{S^2-2ST+T^2-S^2-2ST-T^2}{4} = -\frac{4ST}{4} = -ST.$$Thus$$B^2-A^2=-(-d)^k.$$If $k$ is even, then $(-d)^k=d^k$, so$$B^2-A^2=-d^k,$$equivalently,$$A^2=B^2+d^k.$$Since $x<y$, one has $x-y<0$. For even $k$,$$T=(x-y)^k>0,$$hence$$A=\frac{S+T}{2}>\frac{S-T}{2}=B.$$Thus $B^2<A^2$, consistent with$$A^2=B^2+d^k.$$Since$$S=A+B = \sqrt{A^2}+\sqrt{B^2} = \sqrt{B^2+d^k}+\sqrt{B^2},$$the required identity holds with$$n=B^2.$$If $k$ is odd, then $(-d)^k=-d^k$, and therefore$$B^2-A^2=d^k.$$Hence$$B^2=A^2+d^k.$$Since$$S=A+B = \sqrt{A^2}+\sqrt{B^2} = \sqrt{A^2}+\sqrt{A^2+d^k},$$the required identity holds with$$n=A^2.$$Combining the two parity cases, define$$n= \min!\left{ \left(\frac{S+T}{2}\right)^2, \left(\frac{S-T}{2}\right)^2 \right}.$$The larger of the two numbers equals $n+d^k$, and$$(\sqrt m+\sqrt{m+d})^k = \sqrt n+\sqrt{n+d^k}.$$This completes the proof. ## Verification of Key Steps The integrality argument is obtained from the parity decomposition of the binomial expansions. When $k$ is even, $A$ is an integer and $B=xyN$, so$$A^2,B^2\in\mathbb Z.$$When $k$ is odd,$$A=xM,\qquad B=yN,$$with $M,N\in\mathbb Z$, and therefore$$A^2=mM^2\in\mathbb Z,\qquad B^2=(m+d)N^2\in\mathbb Z.$$The relation between the two radicands comes from$$B^2-A^2=-ST$$together with$$ST=(-d)^k.$$For odd $k$ this yields$$B^2=A^2+d^k.$$For even $k$ this yields$$A^2=B^2+d^k.$$Hence the two radicands always differ by exactly $d^k$. ## Alternative Approaches A more structural proof views $S$ and $T$ as conjugate quantities and uses trace and norm in the quadratic extension generated by $x$ and $y$. The argument above keeps the same construction and derives the needed facts directly from binomial expansions and the identity$$(x+y)^k(x-y)^k=(-d)^k.$$The construction gives an explicit natural number $n$ and verifies directly that$$(\sqrt m+\sqrt{m+d})^k=\sqrt n+\sqrt{n+d^k}.$$