A divisor of (4k²−1)² and the descent hidden in the complementary factor

waddledee42·Nov 9, 2025(edited)

For even k the divisor is obvious, taking

n=k/2

. I need the odd-k obstruction. Direct modular arithmetic doesn't seem to distinguish the cases. Problem: Let

k

be a positive integer. Prove that the number

\left(4 k^{2}-1\right)^{2}

has a positive divisor of the form

8 k n-1

if and only if

k

is even. Source: IMO 2007 Shortlist · MNET-06PT

Open MNET-06PT in the question bank →

6 replies

no_mp_left·Nov 9, 2025(edited)

A stronger lemma is useful: if

4xy-1

divides

(4x^2-1)^2

for positive integers x,y, then x=y. Apply it with

x=k,y=2n

waddledee42·Nov 9, 2025(edited)

That lemma is almost the whole problem. Is there a smaller pair hiding in the quotient?

no_mp_left·Nov 9, 2025(edited)

Yes. If y>x, write the quotient as r. Reducing

r(4xy-1)=(4x^2-1)^2

modulo 4x gives

r\equiv-1

. Thus

r=4xz-1

with z positive. Comparing the quotient with

4x^2-1

gives z<x.

crimsonFlask·Nov 10, 2025

You also need symmetry: modulo

4xy-1

(4y^2-1)^2\equiv16y^4(4x^2-1)^2\equiv0

. So a counterexample (x,y) can be swapped.

waddledee42·Nov 10, 2025

Then minimize x+y among counterexamples. Swap to arrange x<y, replace y by z<x, and get a smaller counterexample. That proves the lemma.

no_mp_left·Nov 10, 2025

Right. Consequently

k=2n

is necessary, and it is sufficient by direct substitution. The descent, not a parity computation alone, is what forces even k.

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A divisor of (4k²−1)² and the descent hidden in the complementary factor

waddledee42·Nov 9, 2025(edited)

For even k the divisor is obvious, taking

n=k/2

. I need the odd-k obstruction. Direct modular arithmetic doesn't seem to distinguish the cases. Problem: Let

k

be a positive integer. Prove that the number

\left(4 k^{2}-1\right)^{2}

has a positive divisor of the form

8 k n-1

if and only if

k

is even. Source: IMO 2007 Shortlist · MNET-06PT

Open MNET-06PT in the question bank →

6 replies

no_mp_left·Nov 9, 2025(edited)

A stronger lemma is useful: if

4xy-1

divides

(4x^2-1)^2

for positive integers x,y, then x=y. Apply it with

x=k,y=2n

waddledee42·Nov 9, 2025(edited)

That lemma is almost the whole problem. Is there a smaller pair hiding in the quotient?

no_mp_left·Nov 9, 2025(edited)

Yes. If y>x, write the quotient as r. Reducing

r(4xy-1)=(4x^2-1)^2

modulo 4x gives

r\equiv-1

. Thus

r=4xz-1

with z positive. Comparing the quotient with

4x^2-1

gives z<x.

crimsonFlask·Nov 10, 2025

You also need symmetry: modulo

4xy-1

(4y^2-1)^2\equiv16y^4(4x^2-1)^2\equiv0

. So a counterexample (x,y) can be swapped.

waddledee42·Nov 10, 2025

Then minimize x+y among counterexamples. Swap to arrange x<y, replace y by z<x, and get a smaller counterexample. That proves the lemma.

no_mp_left·Nov 10, 2025

Right. Consequently

k=2n

is necessary, and it is sufficient by direct substitution. The descent, not a parity computation alone, is what forces even k.