Kvant Math Problem 681
Let
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Solve time: 7m41s
Source on kvant.digital
Problem
- Find integers $a$, $b$, $c$, $d$ such that the numbers $a^2+b^2$, $a^2+b^2+c^2$, $a^2+b^2+c^2+d^2$ are perfect squares.
- Does there exist a sequence consisting of perfect squares such that for any $n$, the sum of $n$ of its first terms is a perfect square?
G. Grigoriev
Exploration
Let
$$S_1=a^2+b^2,\qquad S_2=a^2+b^2+c^2,\qquad S_3=a^2+b^2+c^2+d^2.$$
The first question asks for one chain of three successive square partial sums. A natural attempt is to start with a Pythagorean triple. Taking
$$3^2+4^2=5^2$$
gives the first square. To continue, we need
$$5^2+c^2$$
to be a square. The triple
$$5^2+12^2=13^2$$
works. Then we need
$$13^2+d^2$$
to be a square. The triple
$$13^2+84^2=85^2$$
works. This already solves part 1.
For part 2, write the partial sums as
$$S_n=A_n^2.$$
Since every term of the sequence must itself be a square, we need
$$A_n^2-A_{n-1}^2$$
to be a square for every $n$.
The examples
$$5^2+12^2=13^2,\qquad 13^2+84^2=85^2$$
suggest constructing an infinite chain
$$A_n^2+B_n^2=A_{n+1}^2.$$
The crucial point is to find a recurrence producing infinitely many such linked triples.
Using the identity
$$x^2+\left(\frac{x^2-1}{2}\right)^2 = \left(\frac{x^2+1}{2}\right)^2 \qquad (x\ \text{odd}),$$
if we set
$$A_{n+1}=\frac{A_n^2+1}{2}, \qquad B_n=\frac{A_n^2-1}{2},$$
then
$$A_n^2+B_n^2=A_{n+1}^2.$$
Starting with $A_1=5$ gives
$$5,13,85,\ldots$$
and produces an infinite chain.
Problem Understanding
Both parts are existence problems, so this is a Type D problem.
The first part asks for integers $a,b,c,d$ such that three consecutive partial sums of the squares $a^2,b^2,c^2,d^2$ are perfect squares.
The second part asks whether there exists an infinite sequence of perfect squares whose every initial partial sum is again a perfect square.
The core difficulty is constructing infinitely many squares $A_n^2$ such that each difference
$$A_{n+1}^2-A_n^2$$
is itself a square.
The answer is yes. The identity
$$x^2+\left(\frac{x^2-1}{2}\right)^2 = \left(\frac{x^2+1}{2}\right)^2 \quad (x\ \text{odd})$$
produces a recursive chain of Pythagorean triples.
Proof Architecture
First, construct one finite chain:
$$3^2+4^2=5^2,\qquad 5^2+12^2=13^2,\qquad 13^2+84^2=85^2.$$
This immediately yields suitable $a,b,c,d$.
Next, prove the identity
$$x^2+\left(\frac{x^2-1}{2}\right)^2 = \left(\frac{x^2+1}{2}\right)^2$$
for odd integers $x$.
Then define a sequence $A_1=5$ and
$$A_{n+1}=\frac{A_n^2+1}{2}.$$
Show that every $A_n$ is odd, so the recurrence is well defined in the integers.
Define
$$T_n=\left(\frac{A_n^2-1}{2}\right)^2.$$
Use the identity to prove
$$A_n^2+T_n=A_{n+1}^2.$$
Finally, show inductively that the partial sums of the sequence
$$T_1,T_2,T_3,\ldots$$
are exactly
$$A_2^2,A_3^2,A_4^2,\ldots,$$
hence all are perfect squares.
The step most likely to fail under scrutiny is the verification that the recurrence indeed makes every difference $A_{n+1}^2-A_n^2$ a square.
Solution
For part 1, compute
$$3^2+4^2=25=5^2,$$
$$3^2+4^2+12^2=25+144=169=13^2,$$
and
$$3^2+4^2+12^2+84^2 = 169+7056 = 7225 = 85^2.$$
Hence
$$a=3,\qquad b=4,\qquad c=12,\qquad d=84$$
satisfy the required conditions.
For part 2, consider the identity
$$x^2+\left(\frac{x^2-1}{2}\right)^2 = x^2+\frac{x^4-2x^2+1}{4} = \frac{x^4+2x^2+1}{4} = \left(\frac{x^2+1}{2}\right)^2.$$
Let
$$A_1=5, \qquad A_{n+1}=\frac{A_n^2+1}{2}.$$
Since $A_1$ is odd, and for an odd integer $u$ the number
$$\frac{u^2+1}{2}$$
is again odd, every $A_n$ is an odd integer.
Define
$$T_n=\left(\frac{A_n^2-1}{2}\right)^2.$$
Applying the identity with $x=A_n$ gives
$$A_n^2+T_n=A_{n+1}^2.$$
We claim that
$$T_1+T_2+\cdots+T_n=A_{n+1}^2$$
for every $n\ge1$.
For $n=1$,
$$T_1=A_2^2-A_1^2.$$
Since $A_1=5$ and
$$T_1=\left(\frac{25-1}{2}\right)^2=12^2=144,$$
we obtain
$$T_1=169-25.$$
Thus
$$T_1=A_2^2-A_1^2.$$
Now assume
$$T_1+\cdots+T_n=A_{n+1}^2-A_1^2.$$
Then
$$\begin{aligned} T_1+\cdots+T_n+T_{n+1} &=(A_{n+1}^2-A_1^2)+(A_{n+2}^2-A_{n+1}^2)\ &=A_{n+2}^2-A_1^2. \end{aligned}$$
Hence, for all $n$,
$$T_1+\cdots+T_n=A_{n+1}^2-A_1^2.$$
Since $A_1=5$, append the square $25=A_1^2$ as the first term and consider the sequence
$$25,\ T_1,\ T_2,\ T_3,\ldots .$$
Its first partial sum is
$$25=5^2.$$
For $n\ge2$, the sum of the first $n$ terms equals
$$A_1^2+(T_1+\cdots+T_{n-1}) = A_1^2+(A_n^2-A_1^2) = A_n^2,$$
which is a perfect square.
Thus an infinite sequence of perfect squares with square initial partial sums exists. One such sequence begins
$$25,\ 144,\ 7056,\ 52200624,\ldots$$
corresponding to
$$5,\ 13,\ 85,\ 3613,\ldots$$
for the square roots of the partial sums.
The required construction is
$$\boxed{ 25,\ \left(\frac{5^2-1}{2}\right)^2,\ \left(\frac{13^2-1}{2}\right)^2,\ \left(\frac{85^2-1}{2}\right)^2,\ldots }$$
with
$$A_{n+1}=\frac{A_n^2+1}{2}.$$
Verification of Key Steps
The first delicate step is the identity
$$x^2+\left(\frac{x^2-1}{2}\right)^2 = \left(\frac{x^2+1}{2}\right)^2.$$
Expanding both sides gives
$$x^2+\frac{x^4-2x^2+1}{4} = \frac{x^4+2x^2+1}{4},$$
which matches the square on the right. No hidden number theoretic assumption is involved.
The second delicate step is the integrality of the recurrence. If $A_n$ is odd, then
$$A_n^2\equiv1\pmod 8,$$
so
$$\frac{A_n^2+1}{2}$$
is an integer. Moreover,
$$\frac{1+1}{2}\equiv1\pmod2,$$
so it remains odd. Hence the recurrence never leaves the integers.
The third delicate step is the passage from
$$A_n^2+T_n=A_{n+1}^2$$
to square partial sums. The equality yields
$$T_n=A_{n+1}^2-A_n^2.$$
Summing these differences produces a telescoping series:
$$\sum_{k=1}^{n-1}T_k = A_n^2-A_1^2.$$
Adding the initial term $A_1^2=25$ gives exactly $A_n^2$. Missing this initial square would make the first partial sum equal to $144$ rather than $25$.
Alternative Approaches
The chain
$$5^2+12^2=13^2,\qquad 13^2+84^2=85^2,\qquad 85^2+3612^2=3613^2$$
may also be obtained from the classical parameterization of Pythagorean triples. Setting one parameter equal to $1$ gives
$$x^2+\left(\frac{x^2-1}{2}\right)^2 = \left(\frac{x^2+1}{2}\right)^2,$$
which is exactly the recurrence used above.
Another route is to view the pairs $(A_n,B_n)$ as successive solutions of
$$A_{n+1}^2-A_n^2=B_n^2$$
and derive them from rational points on the unit circle. The recurrence above is preferable because every step is explicit, elementary, and immediately produces the required sequence of square terms and square partial sums.