🧭 Problem (AMC credit)

Let \(M\) be the greatest integer such that both \(M + 1213\) and \(M + 3773\) are perfect squares. What is the units digit of \(M\)?

1️⃣ Assume both are squares

Let

for some positive integers \(P,Q\) with \(Q > P\) (because \(M+3773 > M+1213\)).

Subtract the first from the second:

This is the key: the difference of the two squares is fixed at \(2560\). So we just have to split 2560 into two factors \(Q+P\) and \(Q-P\) of the same parity.

2️⃣ Parity + Maximising \(M\)

From

we get

So \(d\) and \(e\) must be the same parity (both even) to keep \(P,Q\) integral. Since \(2560 = 2^8 \cdot 5\), a natural choice is to take them both even.

We need the greatest \(M\). But

So to maximise \(M\), we maximise \(P\). And to maximise \(P = \tfrac{d-e}{2}\), we want:

• \(d = Q+P\) as large as possible
• \(e = Q-P\) as small as possible

Keep them even and factor \(2560\): the best choice is

This indeed multiplies to \(2560\) and both are even.

Now solve:

3️⃣ Get \(M\)

Recall \(M + 1213 = P^2\). With \(P = 639\):

We don’t actually need the full number to answer the MCQ. We only need the units digit.

Compute the units digit:

• Units digit of \(639^2\) is the same as units digit of \(9^2 = 81\) → units digit \(= 1\).
• So \(M \equiv 1 - 3 \pmod{10}\) (because 1213 ends with 3).
• \(1 - 3 \equiv -2 \equiv 8 \pmod{10}\).

Therefore, units digit of \(M\) is 8.

This matches the official solution method: first make it a difference of squares, then pick \((Q+P, Q-P) = (1280, 2)\).

📝 Quick Quiz

Suppose instead we had \(M+1000 = P^2\) and \(M+2600 = Q^2\). Then \(Q^2 - P^2 = 1600\). Which \((Q+P, Q-P)\) pair below will give the largest possible \(M\)?