Kvant Math Problem 764
The task is purely existential.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m28s
Source on kvant.digital
Problem
Prove that each of the equations
- $x^2+y^3=z^5$;
- $x^2+y^3+z^5=t^7$
has infinitely many solutions in natural numbers.
O. V. Mazurov
Exploration
The task is purely existential. We need infinitely many natural solutions of
$x^2+y^3=z^5,$
and then infinitely many natural solutions of
$x^2+y^3+z^5=t^7.$
A natural first idea is to force all terms to be powers of a common parameter. If
$x=a^{15},\qquad y=b^{10},\qquad z=c^6,$
then the equation becomes
$a^{30}+b^{30}=c^{30}.$
Thus any Pythagorean identity can be raised to the fifteenth power. Taking
$u^2+v^2=w^2,$
we obtain
$u^{30}+v^{30}=w^{30},$
hence
$(u^{15})^2+(v^{10})^3=(w^6)^5.$
This already gives infinitely many solutions of the first equation from the infinite family of Pythagorean triples.
For the second equation, it is tempting to repeat the same idea and reduce everything to an identity of seventh powers. We want
$x^2+y^3+z^5=t^7.$
If we could arrange
$A^7+B^7=C^7,$
that would contradict Fermat. So a different construction is needed.
Using the first equation seems promising. If
$x^2+y^3=z^5,$
then the second equation becomes
$2z^5=t^7.$
To make this hold, choose
$z=2^2s^7,\qquad t=2s^5.$
Indeed,
$2z^5=2(2^2s^7)^5=2^{11}s^{35}=(2s^5)^7=t^7.$
Thus any solution of the first equation with $z=4s^7$ immediately yields a solution of the second equation.
The crucial point is therefore to produce infinitely many solutions of the first equation for which $z$ is divisible by $4$ and is a seventh power times $4$. Since in the Pythagorean construction $z=w^6$, it suffices to choose
$w=(4s^7)^{1/6}.$
That is not generally integral. Instead, choose the Pythagorean parameter itself to be a sixth power:
$w=r^7.$
Then $z=w^6=r^{42}$, and if $r$ is even, say $r=2s$, then
$$\qquad q=2^5s^6.$$
Hence the desired divisibility condition is automatic.
Take the standard triple
$3k,;4k,;5k.$
Then
$$$$
giving
$$$$
Choosing $k=5^6(2s)^7$ makes $5k=(10s)^7$, hence
$z=(10s)^{42}=4q^7.$
This yields infinitely many solutions of the second equation.
Problem Understanding
We must prove the existence of infinitely many natural-number solutions of two Diophantine equations. This is a Type D problem.
The first equation asks for infinitely many triples $(x,y,z)$ satisfying
$x^2+y^3=z^5.$
The second asks for infinitely many quadruples $(x,y,z,t)$ satisfying
$x^2+y^3+z^5=t^7.$
The core difficulty is to construct parametrized families rather than isolated examples. The exponents $2,3,5$ suggest replacing the variables by suitable powers so that all terms become the same exponent. For the second equation, the difficulty is to make the extra term fit a seventh power without invoking an impossible identity of the form $A^7+B^7=C^7$.
The answer is that both equations admit infinite families obtained from Pythagorean identities and a suitable choice of parameters.
Proof Architecture
The first claim is that every Pythagorean triple $u^2+v^2=w^2$ produces a solution of $x^2+y^3=z^5$ via
$x=u^{15},\qquad y=v^{10},\qquad z=w^6,$
because all three terms become thirtieth powers.
The second claim is that the family
$u=3k,\qquad v=4k,\qquad w=5k$
yields infinitely many solutions of the first equation.
The third claim is that if a solution of the first equation has the form
$z=4q^7,$
then
$x^2+y^3+z^5=(2q^5)^7,$
hence it gives a solution of the second equation.
The fourth claim is that by choosing
$k=5^6(2s)^7,$
the resulting value of $z=(5k)^6$ is indeed of the form $4q^7$.
The most delicate step is the last one, where the factorization of $z$ must be checked carefully to ensure that it is exactly $4$ times a seventh power.
Solution
Consider any Pythagorean triple
$u^2+v^2=w^2.$
Raise both sides to the fifteenth power:
$u^{30}+v^{30}=w^{30}.$
Define
$x=u^{15},\qquad y=v^{10},\qquad z=w^6.$
Then
$x^2=u^{30},\qquad y^3=v^{30},\qquad z^5=w^{30},$
and therefore
$x^2+y^3=z^5.$
Thus every Pythagorean triple yields a solution of the first equation.
Now take the infinite family of Pythagorean triples
$$\qquad k\in\mathbb N,$$
since
$(3k)^2+(4k)^2=(5k)^2.$
The corresponding values
$x=(3k)^{15},\qquad y=(4k)^{10},\qquad z=(5k)^6$
satisfy
$x^2+y^3=z^5.$
Different values of $k$ produce different values of $z$, hence infinitely many solutions exist for the first equation.
For the second equation, choose
$$\qquad s\in\mathbb N.$$
Using the solution obtained above, we have
$$=\bigl(5^7(2s)^7\bigr)^6 =(10s)^{42}.$$
Since
$$$$
we may write
$$=4\Bigl(2^5,5^6,s^6\Bigr)^7.$$
Set
$$$$
Then
$$$$
Because $x^2+y^3=z^5$, we obtain
$$$$
Substituting $z=4q^7$ gives
$$=2(4q^7)^5 =2\cdot4^5,q^{35} =2^{11}q^{35} =(2q^5)^7.$$
Define
$$$$
Then
$$$$
As $s$ ranges through the natural numbers, the values of $z=(10s)^{42}$ are distinct, so the resulting quadruples are distinct. Hence infinitely many solutions of the second equation exist.
We have explicitly constructed infinitely many solutions for each equation. This completes the proof.
∎
Verification of Key Steps
The first delicate step is the passage from a Pythagorean triple to a solution of the first equation. Substituting
$$$$
gives
$$$$
The equality
$$$$
holds because it is exactly the fifteenth power of
$$$$
No further algebraic identity is being used.
The second delicate step is proving that $z$ has the form $4q^7$. Starting from
$$$$
one computes
$$$$
Since
$$\qquad 5^{42}=(5^6)^7, \qquad s^{42}=(s^6)^7,$$
it follows that
$$$$
The factor $4$ is essential. Replacing it by a generic statement such as “$z$ is a seventh power up to a constant” would not suffice.
The third delicate step is the computation
$$$$
Using $z=4q^7$,
$$$$
Since
$$$$
choosing
$$$$
gives the required seventh power exactly. Any mistake in the exponent arithmetic would invalidate the construction.
Alternative Approaches
Instead of the family $(3k,4k,5k)$, one may start from the full parametrization of primitive Pythagorean triples,
$$v=2mn,\qquad w=m^2+n^2,$$
and then define
$$$$
This immediately yields infinitely many solutions of the first equation. By imposing suitable congruence and divisibility conditions on $m$ and $n$, one can arrange that $w$ be a multiple of a seventh power and proceed as above.
Another possibility is to search directly for identities among thirtieth powers. Any infinite family
$$$$
coming from a homogeneous quadratic identity produces solutions of the first equation after setting $x=A^{15}$, $y=B^{10}$, $z=C^6$. The Pythagorean construction is preferable because it is completely explicit and requires only the elementary identity $3^2+4^2=5^2$.